Permutation - Combination
SINGLE CHOICE CORRECT QUESTIONS
1.
A road network as shown in the figure connect four cities. In how many ways can you start from any city
(say A) and come back to it without travelling on the same road more than once ?
(A) 8
(B) 12
Ans. (B)
Sol. for
A—B—D—A
A—B—C—D—A
A—B—C—A
for each cities we have three bath.
So for 4 cities will be having 12 path.
Option B is correct.
(C) 9
(D) 16
2.
Boxes numbered 1, 2, 3, 4 and 5 are kept in a row and they are necessarily to be filled with either a red
or a blue ball such that no two adjacent boxes can be filled with blue balls. How many different arrangements
are possible, given that the balls of a given colour are exactly identical in all respects ?
(A) 8
(B) 10
(C) 13
(D) 22
Ans. (C)
Sol.
1 2 3 4 5
Case - 1 all R ---- 1
5!
=5
4!
Case - 3 3R & 2B Þ 6
Case - 4 2R & 3B Þ 1
So total no. of ways = 13
Case - 2
4R & 1B Þ
3.
JPR\COMP.251\Allen(IIT-JEE Wing)\2020–21\Nurture\Mathematics\UNIT - 4
Sameer has to make a telephone call to his friend Harish, Unfortunately he does not remember the 7 digit
phone number. But he remembers that the first three digits are 635 or 674, the number is odd and there
is exactly one 9 in the number. The maximum number of trials that Sameer has to make to be successful
is (A) 10,000
(B) 3402
(C) 3200
(D) 5000
Ans. (B)
Sol. Total digit are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
635 _ _ _ 9 (if last digit in 9)
then total no. trials in 9 × 9 × 9
if 9 is at either tens hundred or thousandth place then total trial
(9 × 9 × 1 × 4) × 3
729 + 81 × 12 = 1701
&
if first three digit is 674 then too same no. trial are there 1701
So total trial are 3402
option B is correct.
1
JEE-Mathematics
4.
The number of ways in which 5 different books can be distributed among 10 people if each person can get at
most one book is 5
10
10
(A) 252
(B) 10
(C) 5
(D) C5 . 5!
Ans. (D)
Sol. 5 different books, 10 people if each person get at most one book.
10
C5 × 5!
Option D is correct.
5.
Number of ways in which 9 different prizes can be given to 5 students, if one particular student receives 4 prizes
and the rest of the students can get any numbers of prizes is 9
10
9
4
5
(A) C4 . 2
(B) C5 . 5
(C) 4 . 4
(D) None of these
Ans. (A)
9
Sol. One particular student receive 4 prizes out of 9 different prizes in C4 ways.
5
rest 4 student may have any no. prizes is 4 .
9
5
9
10
So total no. of ways in C4 × 4 = C4 × 2
Option A is correct.
6.
18 points are indicated on the perimeter of a triangle ABC (see figure).
How many triangles are there with vertices at these points?
(A) 331
(B) 408
(C) 710
(D) 711
Ans. (D)
18
7
Sol.
C3 – 3 . C3
7.
If all the letters of the word “QUEUE” are arranged in all possible manner as they are in a dictionary, then the
rank of the word QUEUE is th
th
th
th
(A) 15
(B) 16
(C) 17
(D) 18
Ans. (C)
4!
3!
Sol. E ---------------;
QE ------------2!
2!
QUEEU ------- 1
QUEUE
th
So rank is 17
Option C is correct.
8.
Number of numbers greater than a million and divisible by 5 which can be formed by using only the digits
1, 2, 1, 2, 0, 5 & 2 is (A) 120
(B) 110
(C) 90
(D) None of these
Ans. (B)
6!
----------- 0 (if unit digit is 0)
3! 2!
& if unit digit 5 then 0 can be either
Tens, Hundredth, Thousandth, Hundredththousandth
-------------------------------- (5)
So no. of numbers be
Sol. 1 ----------
6!
5!
+
´5
3! 2! 3! 2!
6.5.4 5.4
+
´5
2
2
60 + 50 = 110
Option B is correct.
2
0 or 5
JPR\COMP.251\Allen(IIT-JEE Wing)\2020–21\Nurture\Mathematics\UNIT - 4
5!
´5
3!2!
So total no be
1 or 2 or 5
Permutation - Combination
9.
There are 10 red balls of different shades & 9 green balls of identical shades. Then the number of arranging them
in a row so that no two green balls are together is
11
(A) (10 !) . P9
(B) (10 !) .
11
C9
(C) 10 !
Ans. (B)
Sol. First place red balls then in the gap among them place green ball.
11
C9 × 10! × 1
(D) 10 ! 9 !
10.
The sum of all numbers greater than 1000 formed by using the digits 1, 3, 5, 7 such that no digit is being
repeated in any number is (A) 72215
(B) 83911
(C) 106656
(D) 114712
Ans. (C)
Sol. Sum of all no.
(1 + 3 + 5 + 7) (4 – 1)! × 1111 = 106656
Option C is correct.
11.
The number of ways in which 10 boys can take positions about a round table if two particular boys must not be
seated side by side is :
(A) 10 (9) !
(B) 9 (8) !
(C) 7 (8) !
(D) none
Ans. (C)
Sol. 9! – 8! × 2! = 7 . 8!
12.
The number of way in which 10 identical apples can be distributed among 6 children so that each child receives
atleast one apple is (A) 126
(B) 252
(C) 378
(D) None of these
Ans. (A)
Sol.
C1 + C2 + C3 + C4 + C5 + C6 = 10
Þ
n–1
Cr–1 =
10–1
9
C6–1 = C5
= 126
Option A is correct.
JPR\COMP.251\Allen(IIT-JEE Wing)\2020–21\Nurture\Mathematics\UNIT - 4
13.
Number of ways in which 25 identical pens can be distributed among Keshav, Madhav, Mukund and Radhika
such that at least 1, 2, 3 and 4 pens are given to Keshav, Madhav, Mukund and Radhika respectively, is 18
28
24
18
(A) C4
(B) C3
(C) C3
(D) C3
Ans. (D)
Sol.
Let no. pens Keshav has = x1
no. pens Madhav has = x2
no. pens Mukund has = x3
no. pens Radhika has = x4
x1 + x2 + x3 + x4 = 25
But given
x1 = y1 + 1
x2 = y2 + 2
x3 = y3 + 3
x4 = y4 + 4
Þ
y1 + 1 + y2 + 2 + y3 + 3 + y4 + 4 = 25
y1 + y2 + y3 + y4 = 15
15+4–1
18
Þ
C4–1 = C3
option D is correct.
3
JEE-Mathematics
14.
The number of ways in which a mixed double tennis game can be arranged from amongst 9 married couple if no
husband & wife plays in the same game is (A) 756
(B) 3024
(C) 1512
(D) 6048
Ans. (C)
7
9
Sol. Total number of ways = C2 × C2 = 756 ways
Option A is correct
15.
The sum of the digits in the unit’s place of all numbers formed with the help of 3, 4, 5, 6 taken all at a time is
(A) 18
(B) 432
(C) 108
(D) 144
Ans. (C)
Sol. Each digit will appear in units place in 6 numbers
Þ
Total sum = 6(3 + 4 + 5 + 6) = 6(18) = 108
Þ
Correct option is (C)
16.
An old man while dialing a 7 digit telephone number remembers that the first four digits consists of one 1’s, one
2’s and two 3’s. He also remembers that the fifth digit is either a 4 or 5 while has no memorising of the sixth
digit, he remembers that the seventh digit is 9 minus the sixth digit. Maximum number of distinct trials he has to
try to make sure that he dials the correct telephone number, is
(A) 360
(B) 240
(C) 216
(D) none
Ans. (B)
4!
´ 2 ´ 10 = 240
2!
Sol.
17.
The number of different seven digit numbers that can be written using only three digits 1, 2 & 3 under the
condition that the digit 2 occurs exactly twice in each number is (A) 672
(B) 640
(C) 512
(D) None of these
Ans. (A)
Sol. No. of times a perticular digit used
So total no. are
2.
7!
7!
7!
+ 2.
+ 2.
4 !2!
3! 2!2!
5! 2!
7.6.5.4
+ 7.6
2!
210 + 420 + 42 = 672
Option A is correct.
7.6.5 +
4
1
1
2
2
3
4
7!
4! 2!
2
2
3
7!
3!2! 2!
3
2
2
7!
3! 2! 2!
4
2
1
5
2
0
7!
5! 2!
0
2
5
7!
2!5!
7!
2! 4! 1!
JPR\COMP.251\Allen(IIT-JEE Wing)\2020–21\Nurture\Mathematics\UNIT - 4
digit ®
no. of times ®
Permutation - Combination
18.
A 5 digit number divisible by 3 is to be formed using the numerals 0, 1, 2, 3, 4 & 5 without repetition. The total
number of ways this can be done is (A) 3125
(B) 600
(C) 240
(D) 216
Ans. (D)
Sol. 0, 1, 2, 3, 4, 5
sum of all digit 15
5 digit whose sum is divisible by 3 is 1, 2, 3, 4, 5 total five digit no 5!
Second case 0, 1, 2, 4, 5
Total no. = 5! – 4! = 120 – 24 = 96
Total no. = 120 + 96 = 216
Option D is correct.
19.
Number of ways in which 9 different toys can be distributed among 4 children belonging to different age groups in
such a way that distribution among the 3 elder children is even and the youngest one is to receive one toy more is
(A)
(5!)2
8
(B)
9!
2
(C)
9!
3!(2!)3
(D) None of these
Ans. (C)
Sol. So distribution is
9!
3! 2! 2! 2!
3
2
2
2
JPR\COMP.251\Allen(IIT-JEE Wing)\2020–21\Nurture\Mathematics\UNIT - 4
option (C) is correct.
9
5
JEE-Mathematics
SECTION - 1 : MULTIPLE CHOICE CORRECT QUESTIONS
1.
All the 7 digit numbers containing each of the digits 1, 2, 3, 4, 5, 6, 7 exactly once and not divisible by 5 are
arranged in the increasing order. Then th
(A) 1800 number in the list is 3124567
th
(B) 1897 number in the list is 4213567
th
(C) 1994 number in the list is 4312567
th
(D) 2001 number in the list is 4315726
Ans. (BD)
Sol. Given 1, 2, 3, 4, 5, 6, 7
Using these digit exactly one need to make 7 digit no. not divisible by 5.
Þ
unit digit can’nt be 5
No. starting with digit 1
1——————
1 2 3 4 5 5
(no. of choices each place value has)
Þ
600
2— — — — —
600
3 — — — — — 600
th
So 1800 no. will be starting from three & last no. of the series.
Option A is incorrect.
(B)
When first digit is 1 we have 600 no.
1 — — — — — — = 600
2 — — — — — — = 600
3 — — — — — — = 600
41——————
1 × 2 × 3 × 4 × 4 = 96
till now we have 1896
Now 42 — — — — — —
last no. of the 42 series will be 1897 no. 4213567
(D)
2001 no. will from
till 42 — — — — — — series we have 1992 no.
Now 43 — — — — — series has been started now with 4312 — — —
we have 6 more again 3 more no. 4315 — — —
So
last no. is 4315726
Hence Option B & D are correct.
The number of five digit numbers that can be formed using all the digits 0, 1, 3, 6, 8 which are (A) divisible by 4 is 30
(B) divisible by 4 is 60
(C) smaller than 60,000 when digit 8 always appears at ten's place is 6
(D) between 30,000 and 60,000 and divisible by 6 is 18.
Ans. (AD)
Sol. given 0, 1, 3, 6, 8
(A) no. divisible by 4 has last two digit
3! - - - - 08
4 - - - - 16
4 - - - - 36
6
JPR\COMP.251\Allen(IIT-JEE Wing)\2020–21\Nurture\Mathematics\UNIT - 4
2.
Permutation - Combination
6 - - - - 80
6 - - - - 60
4 - - - - 68
total no. = 30
(B) 30000
8 1
3
1 8 0 6
Þ 4 in no.
6 0
0 3
6
Þ 4 in no.
1 8 3 0
8 1
8
Þ 2! 2! = 4
3 6 1 0 6 3 0 1
total no. = 12
Option B in also correct.
(D) 0, 1, 3, 6, 8
3 _ _ _ _ _ 6/8/0
2×3×3
60 _ _ _ _ _
as unit digit has three choices.
then 3 then 2
Þ
total no. = 18
Hence option ABD are correct.
3.
Which of the following statement(s) is/are true :100
(A) C50 is not divisible by 10
(B) n(n – 1)(n – 2) .........(n – r + 1) is always divisible by r! (n Î N and 0 £ r £ n)
(C) Morse telegraph has 5 arms and each arm moves on 6 different positions including the position of rest.
6
Number of different signals that can be transmitted is 5 – 1.
5
(D) There are 5 different books each having 5 copies. Number of different selections is 6 –1.
Ans. (ABD)
100
Sol. (A)
C50
100!
2100 . (1.3.5......99 ) .50!
=
50!50!
50! 50!
Option A is true.
(B)
r !´
n ( n - 1)( n - 2 ) ...... ( n - ( r - 1) ) ( n - r ) !
( n - r )! r !
n
Þ r! × Cr
option B is also correct.
JPR\COMP.251\Allen(IIT-JEE Wing)\2020–21\Nurture\Mathematics\UNIT - 4
(D)
5 different book
C0
C1
B1
C4
C2 C3
B2
B3
B4
B5
5
6 –1
Option D is correct.
Option A, B & D are correct.
7
JEE-Mathematics
4.
A persons wants to invite one or more of his friend for a dinner party. In how many ways can he do so if he has
eight friends : 8
8
2
8
8
8
(A) 2
(B) 2 – 1
(C) 8
(D) C1 + C2 + .....+ C8
Ans. (BD)
Sol. He can invite either 1 or 2 or 3 or 4 --------- or all
So option B & D are correct.
2
5.
3
4
N = 2 . 3 .5 .7, then (A) Number of proper divisors of N(excluding 1 & N) is 118
(B) Number of proper divisors of N(excluding 1 & N) is 120
(C) Number of positive integral solutions of xy = N is 60
(D) Number of positive integral solutions of xy = N is 120
Ans. (AD)
2
3
4
1
Sol. N º 2 . 3 . 5 . 7
Total no. of divisor
= (2 + 1)(3 + 1)(4 + 1)(1 + 1)
= 3 × 4 × 5 × 2 = 120
& no. of proper divison = 118
option A is correct
Now if N = xy
Þ
N can be expressed as a product of two no. in haw many ways.
Which is equal to
1
(120 ) = 60
2
SECTION - 2 : COMPREHENSION BASED QUESTIONS
(SINGLE CHOICE CORRECT QUESTION)
Comprehension – 1
S = {0, 2, 4, 6, 8}. A natural number is said to be divisible by 2 if the digit at the unit place is an even number.
The number is divisible by 5, if the number at the unit place is 0 or 5. If four numbers are selected from S and
a four digit number ABCD is formed.
On the basis of above information, answer the following questions
6.
The number of such numbers which are even (all digits are different) is
(A) 60
(B) 96
(C) 120
Ans. (B)
(D) 204
7.
The number of such numbers which are even (all digits are not different) is
(A) 404
(B) 500
(C) 380
Ans. (A)
(D) None of these
The number of such numbers which are divisible by two and five (all digits are not different) is
(A) 125
(B) 76
(C) 65
(D) 100
Ans. (B)
Sol. Given S º {0, 2, 4, 6, 8}
(6)
— — — —
No. of 4 digit no. using the digit of set S
place values — — — —
no. of ways a digit can be placed at place values 4 × 4 × 3 × 2 = 96
So
(7)
8
option A is correct.
Total no. of 4 digit no. = 4 × 5 × 5 × 5 = 500
total no. of 4 digit no when digit are different = 96
JPR\COMP.251\Allen(IIT-JEE Wing)\2020–21\Nurture\Mathematics\UNIT - 4
8.
Permutation - Combination
So
no. of numbers when digit are not different = 500 – 96 = 404
option A is correct.
(8)
Given no. os divisible by 2 & 5 mean no. is divisible by 10.
So unit digit has to be 0.
Total no. which are divisible
— — — —
4 × 5 × 5 × 1 = 100
total no. which are divisible by 10 & digit are different
— — — —
2 × 3 × 4 × 1 = 24
So
no. of such no. which are divisible by 10 (all digit are not diffferent)
= 100 – 24 = 76
option B is correct.
Comprehension – 2
Let p be a prime number and n be a positive integer, then exponent of p is n! is denoted by E p (n!) and
is given by
énù é n ù é n ù
énù
Ep(n!) = ê ú + ê 2 ú + ê 3 ú + ... + ê k ú
p
p
p
ëp û
ë û ë û ë û
k
k+1
where p < n < p
and [x] denotes the integral part of x.
If we isolate the power of each prime contained in any number N, then N can be written as
N = 2a1 . 3a2 . 5 a3 . 7 a4 ....
where ai are whole numbers.
On the basis of above information, answer the following questions
9.
The exponent of 7 in
(A) 0
Ans. (A)
Sol.
100
C50 is (B) 1
(C) 2
7 100
7 14
7 2
0
100!
50!50!
exponent of 7 in
exponent of 7 in
So
exponent of 7 in
(D) 3
100! = 16
50! = 8
100!
716
= 8 8 =0
50!50!
7 .7
7 100
7 7
7 1
option A is correct.
JPR\COMP.251\Allen(IIT-JEE Wing)\2020–21\Nurture\Mathematics\UNIT - 4
10.
The number of zeros at the end of 108! is (A) 10
(B) 13
Ans. (C)
Sol. No. of zeroes at the end of 108!
will be responsible by no. of 5 & no. of 2
since power of 2 will be more as compared to power 5
(C) 25
(D) 26
5 108
5 21
5 4
0
power of 5 in 108!
= 25
So no. of zeroes at the end 108! = 25
option C is correct.
9
JEE-Mathematics
11.
The exponent of 12 in 100! is (A) 32
(B) 48
Ans. (B)
Sol. exponent of 12 in 100!
(C) 97
(D) None of these
2 100
2 50
2 25
2 12
2
6
2
3
1
power of 2 in 100!
= 97
power of 3 in 100!
= 48
So power of 12
= power of 4
power 4 = 48
So
power of 12 = 48
so
option (B) is correct.
& power 3
& power of 3 = 48
3 100
3 33
3 11
3
3
1
Comprehension – 3
Consider the letters of the word ‘MATHEMATICS’.
12.
Possible number of words taking all letters at a time such that in each word both M’s are together and both T’s
are together but both A’s are not together is
(A)
11!
10!
2!2!2! 2! 2!
8
(B) 7! C2
(C)
6!4 !
2! 2!
(D)
9!
2! 2! 2!
Ans. (B)
Sol. Required number of ways
= (No. of ways when both M and both T are together) – (No. of ways when both M1 both T and both A are
together)
=
9! 8!
9! - 2!8!
=
2! 1
2!
=
8!(7)
æ 8 ´ 7ö
= 7! ´ ç
è 2! ÷ø
2!
8
= 7! × C2 = Option B is correct
13.
Possible number of words in which no two vowels are together is
4!
2!
7! 8
4!
C4
2!
2!
Sol. Total number of ways = C4 ×
7! ´ 4!
2! ´ 2! ´ 2!
Ans. (C)
8
Option ‘C’ is correct
10
(C)
7! 8
4!
C4
2! 2!
2!
(D)
7! 8
4!
C4
2! 2! 2!
2!
JPR\COMP.251\Allen(IIT-JEE Wing)\2020–21\Nurture\Mathematics\UNIT - 4
(B)
8
(A) 7! C 4
Permutation - Combination
SECTION - 1 : NUMERICAL ANSWER BASED QUESTIONS
1.
In how many ways can a team of 6 horses be selected out of a stud of 16, so that there shall always be 3 out of
ABC A'B'C', but never AA', BB' or CC' together ?
Ans. 960
6
Sol. Selecting 3 horses out of ABC A'B'C' is C3 ways
When AA' is always selected among (ABC A'B'C')
4
Remaining (BB'CC') can be selected in C1 ways similarly, when BB' and CC' is selected
\
4
Undesirable ways will be ( C1) × 3
using, total ways–undesirable ways = desired ways we get
6
4
( C3 – ( C1)3) ® This is selection of 3 horses among (ABC A'B'C') under given condition.
10
Remaining 3 can be selected in C3 ways.
6
4
10
Hence, desired ways will be [ C3 – C1 × 3] C3 = 792
2
2
2
Method II : Select one horse each from AA', BB' and CC' hence C1 × C1 × C1 ways. Now select
10
3 horses from remaining 10 horses in C3 ways.
Total ways =
2.
10
2
2
2
C3 × C1× C1 × C1
An examination paper consists of 12 questions divided into parts A & B. Part-A contains 7 questions & Part -B
contains 5 questions. A candidate is required to attempt 8 questions selecting atleast 3 from each part. In how
many maximum ways can the candidate select the questions ?
Ans. 420
Sol. Case I : 5 Q from part A and 3Q from
7
5
B ® C5 × C3 = 21 × 10 = 210
7
5
7
5
Case II : 4 Q from A and 4Q from
B ® C4 × C4 = 175
Case III : 3 Q from A and 5 Q from
B ® C3 × C5 = 35
Total = 210 + 175 + 35 = 420
3.
JPR\COMP.251\Allen(IIT-JEE Wing)\2020–21\Nurture\Mathematics\UNIT - 4
The number of n digit numbers which consists of the digits 1 and 2 only if each digit is to be used at least once,
is equal to 510 then n is ........
Ans. 9
n
Sol. Total number of ways is 2 – 2 = 510
n
Þ
2 = 512
n
9
Þ
2 = (2)
Þ
Value of ‘n’ is 9
4.
5 boys and 4 girls sit in a straight line. If the number of ways in which they can be seated if 2 girls are together
N
and the other 2 are also together but separated from the first 2 are N then
is equal to
100
Ans. 432
st
Sol. Step 1 : Arrange 5 boys in 5! ways
nd
6
Step 2 : Select 2 gaps from 6 gaps for 4 girls (2girls for each gap) in C2 ways.
rd
4
Step 3 : Select 2 girls to sit in one of the gaps and other 2 in remaining selected gaps = C2 ways
st
Step 4 : Arrange 1 , 2 girls in 2! and other 2 in 2! ways
6
4
Hence, total ways ® 5! × C2 × C2 × 2 × 2 = 43200
11
JEE-Mathematics
5.
How many different ways can 15 candy bars be distributed between Ram, Shyam, Ghanshyam and Balram, if
Ram can not have more than 5 candy bars and Shyam must have at least two ? Assume all candy bars to be
alike.
Ans. 440
Sol. Distribute 15 candies among.
Ram (R) + Shyam(S) + Ghanshyam(G) + Balram(B)
with condition given : R+S+G+B=15&R£ 5 & S ³ 2
After giving 2 to Shyam, remaining candies 15–2=13
Now distribute 13 candies in
R, S, G, B in
13 + 4 - 1 16
= C3 ways
13. 3
16
In C3 ways, we have to remove undesirable ways,
when R > 5
Undesirable ways : R > 5 Þ R ³ 6
give at least 6 to R and 2 to S and distribute remaining between R, S, G, B
15 – (2 + 6) = 7 remaining can be distributed between R, S, G, B in =
7 + 4 - 1 10
= C3 ways
7. 4 - 1
10
C3 are the undesirable cases
16
10
Desired ways = C3 – C3 = 440
6.
Find the number of ways in which the letters of the word 'MUNMUN' can be arranged so that no two alike
letters are together.
Ans. 30
Sol. MUNMUN
A : two 'M' are together
B : tow 'U' are together
C : Two 'N' are together
n(A Ç B Ç C) = total – n(AÈBÈC)
én(A) + n(B) + n(C) – n(A Ç B)
ù
6!
= 2! 2!2! – ê –n(B Ç C) – n(C Ç A) + n(A Ç B Ç C)ú
ë
û
=
6!
é 3 ´ 5! 3 ´ 4 !
ù
–ê
–
+ 3!ú
2! 2! 2! ë 2! 2!
2!
û
720
– [90 – 36 + 6]
8
= 90 – 90 + 36 – 6 = 30
=
7.
Find the number of ways in which a selection of 100 balls, can be made out of 100 identical red balls, 100
identical blue balls & 100 identical white balls.
Ans. 5151
total 300 balls out of which 100 balls for be selected.
B + R + W = 100
C2
102
=
8.
(B; R; W ³ 0)
102 ´ 101
= 51 × 101 = 5151
2
There are 5 balls of different colours & 5 boxes of colours same as those of the balls. The number of ways in
which the balls, one in each box could be placed such that exactly one ball goes to the box of its own colour.
Ans. 45
Sol. Total number of ways = Put exactly one ball in its box and then dearrange remaining balls.
12
JPR\COMP.251\Allen(IIT-JEE Wing)\2020–21\Nurture\Mathematics\UNIT - 4
Sol. 100B + 100 R + 100 W
Permutation - Combination
LM
N
1
OP
Q
1 1
1
5
= C1 × 4! 1 - 1! + 2! - 3! + 4! = 5 × 9 = 45
9.
There are four balls of different colours and four boxes of the same colours as those of the balls. The number of
ways in which the balls could be placed one each in a box, such that a ball does not go to a box of its own
colour, is ________.
[JEE 1992]
Ans. 9
Sol. The number of ways in which the ball does not go its own colour box
1 1 1 1ö
æ
= 4! ç 1 - + - + ÷
è 1! 2! 3! 4 ! ø
æ1 1 1 ö
æ 12 - 4 + 1 ö
4! ç - +
÷ = 24 ç
÷=9
2
6
24
24
è
ø
è
ø
SECTION - 2 : MATRIX - MATCH QUESTIONS
Following questions contains statements given in two columns, which have to be matched. The statements in
Column-I are labelled as A, B, C and D while the statements in Column-II are labelled as p, q, r and s. Any given
statement in Column-I can have correct matching with ONE statement in Column-II.
10.
5 balls are to be placed in 3 boxes. Each box can hold all the 5 balls. Number of ways in which the balls can be
placed so that no box remains empty, if :
(A)
(B)
(C)
(D)
Column-I
balls are identical but boxes are different
balls are different but boxes are identical
balls as well as boxes are identical
balls as well as boxes are identical but boxes are kept in a row
Column-II
(p) 2
(q) 25
(r) 50
(s) 6
Ans. (A)-(s); (B)-(q); (C)-(p); (D)-(s)
Sol. (A) 5 ball are identical & 3 boxes are different so no. of ways each box can be filled such that no box remain
empty.
5–1
4
= C3–1 = C2 = 6
option s is correct match.
(B) Here balls are different & box are identical.
So
(C)
So
JPR\COMP.251\Allen(IIT-JEE Wing)\2020–21\Nurture\Mathematics\UNIT - 4
(D)
So
Case-1. 1 . 1 . 3
Þ
Case-2. 1 . 2 . 2
Þ
total no. of ways = 25
Option q is correct match.
Balls as wall as boxes are identical
Case-1 1 . 1 . 3
Case-2 1 . 2 . 2
there are two cases only
Option p is correct match.
5
C1 . 4 C 1
= 10
2!
5
C1 . 4 C 2
= 15
2!
Case-1
1.1.3
1.3.1
3.1.1
Case-2 1 . 2 . 2
2.1.2
2.2.1
total no. of ways 6
option s is correct match.
13
JEE-Mathematics
11.
Consider all the different words that can be formed using the letters of the word HAVANA, taken 4 at a time.
(A)
(B)
(C)
Column-I
Number of such words in which all the 4 letters are different
Number of such words in which there are 2 alike letters &
2 different letters.
Number of such words in which A's never appear together
Column-II
(p)
36
(q)
42
(r)
(s)
37
24
Ans. (A)-(s); (B)-(p); (C)-(q)
Sol. (A) H A V N
A
A
4
all four letter are different C4 = 1! = 24
option s is correct match.
3
(B) 2 alike letters can be taken is one way & 2 different letters can be taken in C2
3
so total no. of wards = C 2 ´
(C)
4!
= 3 × 4 . 3 = 36
2!
option p is correct.
A’s never appear together total no. of words.
4!
= 12
3!
4!
3
Case-2 2 A’s & 2 diff. = C 2 . = 36
2!
Case 1 A’s & 3 diff. = 4! = 24
total no. of words = 72
3
Now is case 1 when all three A’s are together = C1 × 2! AAAD
in case-1 when 2 A’s are together AAD A
3
Þ
C1 × 2!
in case - 2 when 2 A’s are together
3
Þ
C2 × 3!
So
total no. of words when in case - 1 & case - 2 A’s are together
3
3
3
= C2 × 2! + C2 × 2! + C2 × 3! = 12 + 12 + 18 = 30
No. of words in which A’s are never together.
= 72 – 30 = 42
option q is correct match.
3
Case-1 3 A’s 1 diff. = C1 ´
Match the column
Column-I
(A)
(B)
Column-II
21
22
23
24
20
20
C2 + C2 + C2 + C2 + C2 + C3 is equal to
In the adjoining figure number of progressive ways to reach from
(0, 0) to (4, 4) passing through point (2, 2) are
(particle can move on horizontal or vertical line)
(p)
(q)
102
2300
(r)
82
(4,4)
4
3
(2,2)
2
1
0
(C)
14
1
2
3
4
The number of 4 digit numbers that can be made with the digits
JPR\COMP.251\Allen(IIT-JEE Wing)\2020–21\Nurture\Mathematics\UNIT - 4
12.
Permutation - Combination
1, 2, 3, 4, 3, 2
(D)
ì 500! ü
If í k ý = 0 , then the maximum natural value of k is equal to
î 14 þ
(s)
36
(where {.} is fractional part function)
Ans. (A)-(q); (B)-(s); (C)-(p); (D)-(r)
24
23
22
21
20
20
Sol. (A)
C2 + C2 + C2 + C2 + C2 + C3
n
n
n+1
Use
Cr + Cr–1 =
Cr
25
C3 =
25!
25 ´ 24 ´ 23
=
= 2300
3! 22!
6
option q is correct match.
(B)
4!
No. of way to move from A ® B
2! 2!
&
no. of ways from B ® C
C
(4,4)
4!
2! 2!
B
(2,2)
4!
4!
total no of ways 2! 2! . 2! 2! = 36
option s in correct match.
(C) 1 2 3 4
2 3
Case-1 When all four digit are different
4! = 24
Case-2 When 2 are alike & 2 are different
A
4!
= 72
2!
Case-3 When 2 are alike of one kind & 2 are of other kind
2
C1 ´ 3 C 2 ´
2
Total words
option p is correct match.
(D)
C2 ´
4!
=6
2! 2!
= 102
500!
14 k
expenent of 7 in 500!
= 82
So power of 14 is also
7 500
7 71
7 10
1
= 82
option r is correct match.
JPR\COMP.251\Allen(IIT-JEE Wing)\2020–21\Nurture\Mathematics\UNIT - 4
So
15
JEE-Mathematics
SECTION - 1 : SINGLE CHOICE CORRECT QUESTIONS
1.
From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary are to be selected and arranged
in a row on a shelf so that the dictionary is always in the middle. Then the number of such arrangements
is :[AIEEE 2009]
(1) At least 750 but less than 1000
(2) At least 1000
(3) Less than 500
(4) At least 500 but less than 750
Ans. (2)
6
3
Sol. C4 × C1 × 4! × 1 .......dictionaries.....
So Option (2) is correct
2.
Statement - 1 – The number of ways of distributing 10 identical balls in 4 distinct boxes such that no box
9
is empty is C3.
[AIEEE-2011]
9
Statement - 2 – The number of ways of choosing any 3 places from 9 different places is C3.
(1) Statement-1 is true, Statement-2 is false.
(2) Statement-1 is false, Statement-2 is true
(3) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1
(4) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
Ans. (3)
Sol. B1 + B2 + B3 + B4 = 10. Identical balls
10 + 4 – 1
9
C4 – 1 = C3
Statement-1 is correct
Statment-2 is also correct
Option (3) is correct
3.
There are 10 points in a plane, out of these 6 are collinear. If N is the number of triangles formed by joining
these points, then :
[AIEEE-2011]
(1) N > 190
(2) N < 100
(3) 100 < N < 140
(4) 140 < N < 190
Ans. (2)
Sol. Number of triangle is
6
4
6
4
4
C1 . C2 + C2 . C1 + C3
6 × 6 + 15 × 4 + 4
36 + 60 + 4 = 600
4.
G5
G4
G6
G3
G2
G7
16
G1
JPR\COMP.251\Allen(IIT-JEE Wing)\2020–21\Nurture\Mathematics\UNIT - 4
If seven women and seven men are to be seated around a circular table such that there is a man on either
side of every woman, then the number of seating arrangements is :
[AIEEE ONLINE-2012]
2
2
(1) 7!
(2) 6! 7!
(3) (6!)
(4) (7!)
Ans. (2)
Sol. Total ways 7! × 6! (Circular Arrangement)
Permutation - Combination
m
5.
n
If n = C2, then the value of C2 is given by :
m+2
m–1
(1) 2( C4)
(2)
C4
Ans. (4)
Sol.
n
C2 =
(3)
m+1
C4
[AIEEE ONLINE-2012]
m+1
(4) 3( C4)
n(n - 1)
1× 2
m(m - 1) é m(m - 1) ù
- 1ú
1 × 2 êë 1 × 2
û
=
1× 2
=
m(m - 1)(m 2 - m - 2)
8
(m + 1)m(m - 1)(m - 2)
24
m+1
=3×
C4.
= 3×
6.
The number of arrangements that can be formed from the letters a, b, c, d, e, f, taken 3 at a time without
repetition and each arrangement containing at least one vowel, is :
[AIEEE ONLINE-2012]
(1) 72
(2) 96
(3) 24
(4) 128
Ans. (2)
Sol. a, e
b, c, d, f
2
4
C1 × C2 × 3! +
2
4
C2 × C1 × 3!
= 12 × 6 + 4 × 6
= 96
7.
A committe of 4 persons is to be formed from 2 ladies, 2 old men and 4 young men such that it includes
at least 1 lady, at least 1 old man and at most 2 young men. Then the total number of ways in which this
committe can be formed is :
[JEE-MAIN ONLINE-2013]
(1) 41
(2) 40
(3) 16
(4) 32
Ans. (1)
Sol. Ladies
oldmen
young men
2
2
4
2
C
4
2
2
4
C1 × 2 1 × C2 + C1 × C2 × C1 +
2
2
4
2
2
C2 × C1 × C1 + C2 × C2
= 24 + 8 + 8 + 1
= 41
JPR\COMP.251\Allen(IIT-JEE Wing)\2020–21\Nurture\Mathematics\UNIT - 4
8.
The number of ways in which an examiner can assign 30 marks to 8 questions, giving not less than 2 marks
to any question, is :
[JEE-MAIN ONLINE-2013]
30
21
21
30
(1) C8
(2) C7
(3) C8
(4) C7
Ans. (2)
Sol. Q1 + Q2 + ....... + Q8 = 30
Qi ³ 2
Þ
" i = 1, 2, ....., 8
Q1 + Q2 + ...... + Q8 = 14
Qi ³ 0
14 + 8 – 1
=
21
" i = 1, 2, ..... , 8
C8 – 1
C7
17
JEE-Mathematics
9.
5 - digit numbers are to be formed using 2, 3, 5, 7, 9 without repeating the digits. If p be the number of such
numbers that exceed 20000 and q be the number of those that lie between 30000 and 90000, then p : q is
:
[JEE-MAIN ONLINE-2013]
(1) 6 : 5
(2) 4 : 3
(3) 5 : 3
(4) 3 : 2
Ans. (3)
Sol. p = 5! = 120
3
q = C1 × 4! = 72 (digit 3, 5 or 7)
p
120
5
=
=
q
72
3
10.
Let A and B be two sets containing 2 elements and 4 elements respectively. The number of subsets of
A × B having 3 or more elements is
[JEE (Main)-2013]
(1) 256
(2) 220
(3) 219
(4) 211
Ans. (3)
Sol. Let A º {a1 a2]
B º {b1, b2, b3, b4}
A × B = {(a1 b1) (a1 b2) (a1 b3) (a1 b4) (a2 b1) (a2 b2) (a2 b3) (a2 b4)}
So
number of subset containing 3 or more element from A × B.
8
8
8
8
8
8
= C3 + C4 + C5 + C6 + C7 + C8
8
8
8
8
= 2 – C2 – C1 – C0
= 256 – 28 – 8 – 1
= 219
11.
Let Tn be the number of all possible triangles formed by joining vertices of an n-sided regular polygon. If
Tn+1 – Tn = 10, then the value of n is :
[JEE (Main)-2013]
(1) 7
(2) 5
(3) 10
(4) 8
Ans. (2)
n+1
n
Sol.
C3 – C3 = 10
(n + 1)!
n!
= 10
3!(n - 2)! 3!(n - 3)!
(n + 1) n(n – 1) –n(n – 1) (n – 2) = 60
n(n – 1) [n + 1 – n + 2] = 60
Þ n(n – 1) = 20
2
Þ n – n – 20 = 0
Þ (n + 4) (n – 5) = 0
n=5
Option (2) is correct
The sum of the digits in the unit's place of all the 4-digit numbers formed by using the number 3, 4, 5 and
6, without repetition, is :
[JEE-MAIN ONLINE-2014]
(1) 432
(2) 36
(3) 18
(4) 108
Ans. (4)
Sol. Each digit will appear 6 times at unit place so
6(3 + 4 + 5 + 6) = 108
18
JPR\COMP.251\Allen(IIT-JEE Wing)\2020–21\Nurture\Mathematics\UNIT - 4
12.
Permutation - Combination
13.
An eight digit number divisible by 9 is to be formed using digits from 0 to 9 without repeating the digits. The
number of ways in which this can be done is :
[JEE-MAIN ONLINE-2014]
(1) 18 (7!)
(2) 40 (7!)
(3) 36 (7!)
(4) 72 (7!)
Ans. (3)
Sol. 4[7 × 7!] + 8!
('0' included) ('0' excluded)
= (28 + 8) 7!
= 36 × 7!
14.
Two women and some men participated in a chess tournament in which every participant played two games
with each of the other participants. If the number of games that the men played between them- selves exceeds
the number of games that the men played with the women by 66, then the number of men who participated
in the tournament lies in the interval :
[JEE-MAIN ONLINE-2014]
(1) (14, 17)
(2) [8, 9]
(3) (11, 13]
(4) [10, 12)
Ans. (4)
Sol. Men
Women
n
2
n
n
2
2( C2 – C1 × C1) = 66
2
n – 5n – 66 = 0 n = 11
15.
The number of ways of selecting 15 teams from 15 men and 15 women, such that each team consists of a
man and a woman, is :
[JEE-MAIN ONLINE-2015]
2
(1) 15!
(2) 14!
(3) (15!)
(4) None of these
Ans. (1)
16.
If in a regular polygon the number of diagonals is 54, then the number of sides of this polygon is :
[JEE-MAIN ONLINE-2015]
(1) 10
(2) 9
(3) 6
(4) 12
Ans. (4)
Sol.
n(n - 3)
= 54
2
n = 12
17.
JPR\COMP.251\Allen(IIT-JEE Wing)\2020–21\Nurture\Mathematics\UNIT - 4
The number of integers greater than 6,000 that can be formed, using the digits 3, 5, 6, 7 and 8, without
repetition, is :
[JEE (Main)-2015]
(1) 216
(2) 192
(3) 120
(4) 72
Ans. (2)
Sol. 3, 5, 6, 7, 8
(i) 4 digit =3 × 4 × 3 × 2 = 72
(ii) 5 digit = 120
Total = 192
18.
If all the words (with or without meaning) having five letters, formed using the letters of the word SMALL and
arranged as in a dictionary: then the position of the word SMALL is :
[JEE (Main)-2016]
(1) 46th
(2) 59th
(3) 52nd
(4) 58th
Ans. (4)
19
JEE-Mathematics
Sol.
A LLMS ®
4!
= 12 ;
2!
L A L M S ® 4! = 24
M ALLS ®
4!
= 12 ;
2!
SA LLM ®
S L A L M ® 3! = 6;
3!
=3
2!
SM ALL®1
Hence position is 58.
19.
If all the words, with or without meaning, are written using the letters of the word QUEEN and are arranged
as in English dictionary, then the position of the word QUEEN is :
[JEE-MAIN ONLINE-2017]
th
th
th
th
(1) 44
(2) 46
(3) 45
(4) 47
Ans. (2)
Sol. E ................. 4! = 24
4!
= 72
2!
QE ............... 3! = 6
N ................
3!
= 3
2!
QUEEN = 1 = 1
th
= 46
QN ..............
20.
The number of ways in which 5 boys and 3 girls can be seated on a round table if a particular boy B 1 and
a particular girl G1 never sit adjacent to each other, is :
[JEE-MAIN ONLINE-2017]
(1) 7!
(2) 5 × 7!
(3) 5 × 6!
(4) 6 × 6!
Ans. (3)
Sol. Total ways = 7!
B, G, together = 6! 2!
Never together = 7! – 6! 2!
= 5 × 6!
21.
From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary are to be selected and arranged
in a row on a shelf so that the dictionary is always in the middle. The number of such arrangements is :
[JEE-MAIN-2018]
(1) at least 1000
(2) less than 500
(3) at least 500 but less than 750
(4) at least 750 but less than 100
Ans. (1)
Sol. Ways = 6 C 4 × 3 C1 ´ 4!
6´5
´ 3 ´ 24
2
N1 N2 P1 N3 N4 = 1080
22.
Consider a class of 5 girls and 7 boys. The number of different teams consisting of 2 girls and 3 boys that
can be formed from this class, if there are two specific boys A and B, who refuse to be the members of the
same team, is:
[JEE-MAIN ONLINE-2019]
(1) 200
(2) 300
(3) 500
(4) 350
20
JPR\COMP.251\Allen(IIT-JEE Wing)\2020–21\Nurture\Mathematics\UNIT - 4
=
Permutation - Combination
Ans. (2)
Sol. Required number of ways
= Total number of ways – When A and B are always included.
= 5 C 2 . 7 C 3 - 5 C15 C 2 = 300
23.
The number of natural numbers less than 7,000 which can be formed by using the digits 0, 1, 3, 7, 9 (repitition
of digits allowed) is equal to :
[JEE-MAIN ONLINE-2019]
(1) 250
(2) 374
(3) 372
(4) 375
Ans. (2)
Sol.
a1 a 2
a3
3
Number of numbers = 5 – 1
a4
a1 a 2
a3
2 ways for a4
3
Number of numbers = 2 × 5
3
3
Required number = 5 + 2 × 5 – 1
= 374
24.
Let S = {1,2,3, ...., 100}. The number of non-empty subsets A of S such that the product of elements in
A is even is :
[JEE-MAIN ONLINE-2019]
50 50
100
50
50
(1) 2 (2 –1)
(2) 2 –1
(3) 2 –1
(4) 2 +1
Ans. (1)
Sol. S = {1,2,3------100}
= Total non empty subsets-subsets with product of element is odd
100
50
= 2 –1–1[(2 –1)]
100
50
=2 –2
50 50
= 2 (2 –1)
25.
There are m men and two women participating in a chess tournament. Each participant plays two games with
every other participant. If the number of games played by the men between themselves exceeds the number
of games played between the men and the women by 84, then the value of m is :
[JEE-MAIN ONLINE-2019]
(1) 9
(2) 11
(3) 12
(4) 7
Ans (3)
Sol. Let m-men, 2-women
m
m
2
C2 × 2 = C1 C1 . 2 + 84
2
m – 5m – 84 = 0
Þ
(m – 12) (m + 7) = 0
m = 12
JPR\COMP.251\Allen(IIT-JEE Wing)\2020–21\Nurture\Mathematics\UNIT - 4
26.
The sum of all natural numbers 'n' such that 100 < n < 200 and H.C.F. (91, n) > 1 is :
[JEE-MAIN ONLINE2019]
(1) 3221
(2) 3121
(3) 3203
(4) 3303
Ans. (1)
Sol. (105 + 112 + 119 .... + 196) + (104 + 117 + 130 + .... 195) – (182) = 3221
27.
All possible numbers are formed using the digits 1, 1, 2, 2, 2, 2, 3, 4, 4 taken all at a time. The number of such
numbers in which the odd digits occupy even places is :
[JEE-MAIN ONLINE2019]
(1) 175
(2) 162
(3) 160
(4*) 180
Ans. (4)
21
JEE-Mathematics
(odd digits)
4p3
´
2!
Sol.
(even digits)
6!
4! 2!
= 12 × 15 = 180
28.
A committee of 11 members is to be formed from 8 males and 5 females. If m is the number of ways the
committee is formed with at least 6 males and n is the number of ways the committee is formed with at least
3 females, then :
[JEE-MAIN O NLINE2019]
(1*) m = n = 78
(2) n = m – 8
(3) m + n = 68
(4) m = n = 68
Ans. (1)
sol. 8 males, 5 females
8
5
8
5
8
5
m = C6 . C5 + C7 . C4 + C8 . C3
m = 28 + 40 + 10 = 78
5
8
5
8
5
8
n = C3 . C8 + C4 . C7 + C5 . C6
n = 78
29.
A group of students comprises of 5 boys and n girls. If the number of ways, in which a team of 3 students
can randomly be selected from this group such that there is at least one boy and at least one girl in each team,
is 1750, then n is equal to :
[JEE-MAIN ONLINE-2019]
(1*) 25
(2) 28
(3) 27
(4) 24
Ans. (1)
Sol. 5 boys, n girls
ways of selection of 3 including atleast one boy are girls is
5
Þ
Þ
30.
C1 . n C 2 + 5 C 2 . n C1 = 1750
n(n - 1)
.5 + n.(10) = 1750
2
5n
[(n - 1) + 4] = 1750
2
n(n + 3) = 700 = 25 × 28
n = 25
Total number of 6-digit numbers in which only and all the five digits 1, 3, 5, 7 and 9 appear, is :
[JEE-MAIN ONLINE2020]
5
1
6
(6!)
(2) 5
(3) (6!)
(4) 6!
2
2
NTA Ans. (1)
ALLEN Ans. (1)
Sol. Total number of 6-digit numbers in which only and all the five digits 1, 3, 5, 7 and 9 is
6!
5
C1 ´
2!
th
31.
If the number of five digit numbers with distinct digits and 2 at the 10 place is 336 k, then k is equal to :
[JEE-MAIN ONLINE2020]
(1) 8
(2) 6
(3) 4
(4) 7
NTA Ans. (1)
ALLEN Ans. (1)
22
JPR\COMP.251\Allen(IIT-JEE Wing)\2020–21\Nurture\Mathematics\UNIT - 4
(1)
Permutation - Combination
Sol. _ _ _ 2 _
No. of five digits numbers =
No. of ways of filling remaining 4 places
=8×8×7×6
k=
\
8 ´8 ´7´6
=8
336
32.
Let n > 2 be an integer. Suppose that there are n Metro stations in a city located along a circular path. Each
pair of stations is connected by a straight track only. Further, each pair of nearest stations is connected by blue
line, whereas all remaining pairs of stations are connected by red line. If the number of red lines is 99 times
the number of blue lines, then the value of n is :[JEE MAINS - ONLINE - 2020]
(1) 199
(2) 101
(3) 201
(4) 200
Official Ans. by NTA (3)
1
Sol. Number of blue lines = Number of sides = n
n
2
Number of red lines = number of diagonals
n
= C2 – n
n
C2 – n = 99 n Þ
n(n - 1)
- n = 99 n
2
3
n -1
- 1 = 99 Þ n = 201
2
4
33.
There are 3 sections in a question paper and each section contains 5 questions. A candidate has to answer
a total of 5 questions, choosing at least one question from each section. Then the number of ways, in which
the candidate can choose the questions, is :
[JEE MAINS - ONLINE - 2020]
(1) 1500
(2) 2255
(3) 3000
(4) 2250
Official Ans. by NTA (4)
Sol. A
B
C
5
5
1
2
2
1
1
3
JPR\COMP.251\Allen(IIT-JEE Wing)\2020–21\Nurture\Mathematics\UNIT - 4
34.
5
2
2
1
2
2
1
1
3
3
1
1
1
Total number of selection
5
5
5
5
5
5
= ( C1 C2 C2)·3 + ( C1 C1 C3)·3
= 5 · 10 · 10 · 3 + 5 · 5 · 10 · 3
= 2250
A scientific committee is to be formed from 6 Indians and 8 foreigners, which includes at least 2 Indians and
double the number of foreigners as Indians. Then the number of ways, the committee can be formed, is :
(1) 1625
(2) 575
Official Ans. by NTA (1)
Sol.
Indians Foreigners
2
4
3
6
4
8
(3) 560
[JEE MAINS - ONLINE - 2021]
(4) 1050
Number of ways
6
C 2 ´ 8 C 4 = 1050
6
C 3 ´ 8 C 6 = 560
6
C 4 ´ 8 C 8 = 15
23
JEE-Mathematics
Total number of ways = 1625
35.
The total number of positive integral solutions (x, y, z) such that xyz = 24 is :
(1) 36
(2) 24
Official Ans. by NTA (4)
3
1
Sol. xyz = 2 × 3
(3) 45
[JEE MAINS - ONLINE - 2021]
(4) 30
Let x = 2a1 ´ 3b1
y = 2a 2 ´ 3b2
z = 2a 3 ´ 3b2
Now a1 + a2 + a3 = 3.
5
No. of non-negative intergal sol = C2 = 10
& b1 + b2 + b3 = 1
n
3
No. of non-negative intergal sol = C2 = 3
Total ways = 10 × 3 = 30.
36.
The number of seven digit integers with sum of the digits equal to 10 and formed by using the digits 1,2 and
3 only is
[JEE MAINS - ONLINE - 2021]
(1) 42
(2) 82
(3) 77
(4) 35
Official Ans. by NTA (3)
Sol. (I) First possiblity is 1, 1, 1, 1, 1, 2, 3
7!
= 7 × 6 = 42
5!
(II) Second possiblity is 1, 1, 1, 1, 2, 2, 2
required number =
7!
7´6´5
= 35
required number = 4! 3! =
6
Total = 42 + 35 = 77
37.
38.
If the sides AB, BC and CA of a triangle ABC have 3, 5 and 6 interior points respectively, then the total number
of triangles that can be constructed using these points as vertices, is equal to :
(1) 364
(2) 240
Official Ans. by NTA (3)
24
(3) 333
[JEE MAINS - ONLINE - 2021]
(4) 360
JPR\COMP.251\Allen(IIT-JEE Wing)\2020–21\Nurture\Mathematics\UNIT - 4
Team 'A' consists of 7 boys and n girls and Team 'B' has 4 boys and 6 girls. If a total of 52 single matches
can be arranged between these two teams when a boy plays against a boy and a girl plays against a girl, then
n is equal to :
[JEE MAINS - ONLINE - 2021]
(1) 5
(2) 2
(3) 4
(4) 6
Official Ans. by NTA (3)
Sol. Total matches between boys of both team
7
4
= C1 × C1 = 28
Total matches between girls of both
n
6
team = C1 C1 = 6n
Now, 28 + 6n = 52
Þ n=4
Permutation - Combination
Sol.
Total Number of triangles formed
14
3
5
6
= C3 – C3 – C3 – C3
= 333
39.
Option (3)
The sum of all the 4-digit distinct numbers that can be formed with the digits 1, 2, 2 and 3 is:
(1) 26664
(2) 122664
Official Ans. by NTA (1)
Sol. Digits are 1, 2, 2, 3
(3) 122234
[JEE MAINS - ONLINE - 2021]
(4) 22264
4!
= 12 .
2!
total numbers when 1 at unit place is 3.
2 at unit place is 6
3 at unit place is 3.
3
2
So, sum = (3 + 12 + 9) (10 + 10 + 10 + 1)
= (1111) × 24
= 26664
total distinct numbers
40.
If n Pr = n Pr +1 and
n
(1) 1
Official Ans. by NTA (3)
Sol.
n
Þ
(2) 4
(3) 2
[JEE MAINS - ONLINE - 2021]
(4) 3
n!
n!
=
(n - r)! (n - r - 1)!
(n – r) = 1
n
Cr = Cr–1
n
Þ
n!
n!
=
r !(n - r)! (r - 1)!(n - r + 1)!
Þ
1
1
=
r(n - r)! (n - r + 1)(n - r)!
Þ
Þ
JPR\COMP.251\Allen(IIT-JEE Wing)\2020–21\Nurture\Mathematics\UNIT - 4
n
Pr = Pr+1 Þ
C r = n Cr -1 , then the value of r is equal to:
n–r+1=r
n + 1 = 2r
(1) Þ 2r – 1 – r = 1 Þ r = 2
…(1)
…(2)
SECTION - 2 : NUMERICAL ANSWER BASED QUESTIONS
1.
An urn contains 5 red marbles, 4 black marbles and 3 white marbles. Then the number of ways in which 4
marbles can be drawn so that at the most three of them are red is ....
[JEE-MAIN ONLINE2020]
NTA Ans. (490.00)
25
JEE-Mathematics
ALLEN Ans. (490.00 OR 13.00)
Sol. The question does not mention that whether same coloured marbles are distinct or identical.
12
5
So, assuming they are distinct our required answer = C4 – C4 = 490
And, if same coloured marbles are identical then required answer = (2 + 3 + 4 + 4) = 13
2.
If the letters of the word 'MOTHER' be permuted and all the words so formed (with or without meaning) be
listed as in a dictionary, then the position of the word 'MOTHER' is ______. [JEE MAINS - ONLINE - 2020]
Official Ans. by NTA (309.00)
Sol. MOTHER
1®E
2®H
3®M
4®O
5®R
6®T
So position of word MOTHER in dictionary
2 × 5! + 2 × 4! + 3 × 3! + 2! + 1
= 240 + 48 + 18 + 2 + 1
= 309
3.
The total number of 3-digit numbers, whose sum of digits is 10, is _____. [JEE MAINS - ONLINE - 2020]
Official Ans. by NTA (54)
Sol. Let three digit number is xyz
x + y + z = 10 ; x ³ 1, y ³ 0 z ³ 0 ..... (1)
Let T = x – 1 Þ x = T + 1 where T ³ 0
Put in (1)
T+y+z=9;
0 £ T £ 8, 0 £ y, z £ 9
No. of non negative integral solution
= 9+3–1C3–1 – 1 (when T = 9)
= 55 – 1 = 54
4.
A test consists of 6 multiple choice questions, each having 4 alternative answers of which only one is correct.
The number of ways, in which a candidate answers all six questions such that exactly four of the answers are
correct, is ____ .
[JEE MAINS - ONLINE - 2020]
Official Ans. by NTA (135)
Sol. Ways = 6C4 · 14 · 32
= 15 × 9
= 135
The number of words, with or without meaning, that can be formed by taking 4 letters at a time from the
letters of the word 'SYLLABUS' such that two letters are distinct and two letters are alike, is ________.
[JEE MAINS - ONLINE - 2020]
Official Ans. by NTA (240)
Sol. S2YL2ABU
ABCC type words
=
2
C ´ 5 C2
{1
{
selection of
two alike
letters
= 240
26
selection of
of two
distinct letters
´
4
2
{
arrangement of
selected letters
JPR\COMP.251\Allen(IIT-JEE Wing)\2020–21\Nurture\Mathematics\UNIT - 4
5.
Permutation - Combination
6.
The number of words (with or without meaning) that can be formed from all the letters of the word "LETTER"
in which vowels never come together is ________.
[JEE MAINS - ONLINE - 2020]
Official Ans. by NTA (120.00)
Sol. LETTER
vowels = EE, consonant = LTTR
_L_T_ T_ R_
4! 5
2!
´ C 2 ´ = 12 ´ 10 = 120
2!
2!
7.
The students S1, S2,....., S10 are to be divided into 3 groups A, B and C such that each group has at least
one student and the group C has at most 3 students. Then the total number of possibilities of forming such
groups is _____.
[JEE MAINS - ONLINE - 2021]
Official Ans. by NTA (31650)
Sol. If group C has one student then number of groups
10
9
C1[2 – 2] = 5100
If group C has two students then number of groups
10
8
C2[2 – 2] = 11430
If group C has three students then number of groups
10
7
= C3 × [2 – 2] = 15120
So total groups = 31650
8.
The total number of numbers, lying between 100 and 1000 that can be formed with the digits 1, 2, 3, 4, 5,
if the repetition of digits is not allowed and numbers are divisible by either 3 or 5, is ______.
[JEE MAINS - ONLINE - 2021]
Official Ans. by NTA (32)
Sol. We need three digits numbers.
Since 1 + 2 + 3 + 4 + 5 = 15
So, number of possible triplets for multiple of 15 is 1 × 2 × 2
so Ans. = 4 ´ 3 + 4 ´ 3 - 1 ´ 2 ´ 2 = 32
9.
n
n
The total number of two digit numbers 'n', such that 3 + 7 is a multiple of 10, is ______.
[JEE MAINS - ONLINE - 2021]
Official Ans. by NTA (45)
n
n
Sol. for 3 + 7 to be divisible by 10
n can be any odd number
\ Number of odd two digit numbers = 45
10.
The total number of 4-digit numbers whose greatest common divisor with 18 is 3, is ______.
JPR\COMP.251\Allen(IIT-JEE Wing)\2020–21\Nurture\Mathematics\UNIT - 4
[JEE MAINS - ONLINE - 2021]
Official Ans. by NTA (1000)
Sol. Let N be the four digit number
gcd(N,18) = 3
Hence N is an odd integer which is divisible by 3 but not by 9.
4 digit odd multiples of 3
1005, 1011,......., 9999 ® 1500
4 digit odd multiples of 9
1017, 1035,......., 9999 ® 500
Hence number of such N = 1000
27
JEE-Mathematics
11.
The number of times the digit 3 will be written when listing the integers from 1 to 1000 is
[JEE MAINS - ONLINE - 2021]
Official Ans. by NTA (300)
Sol. 3_ _ = 10 × 10 = 100
_3 _ = 10 × 10 = 100
_ _ 3 = 10 × 10 =
100
300
12.
There are 15 players in a cricket team, out of which 6 are bowlers, 7 are batsmen and 2 are wicketkeepers. The
number of ways, a team of 11 players be selected from them so as to include at least 4 bowlers, 5 batsmen and
1 wicketkeeper, is__________.
[JEE MAINS - ONLINE - 2021]
Official Ans. by NTA (777)
Sol. 15 : Players
6 : Bowlers
7 : Batsman
2 : Wicket keepers
Total number of ways for :
at least 4 bowlers, 5 batsman & 1 wicket keeper
6
7
2
7
2
6
7
2
= C4( C6 × C1 + C5 × C2) + C5 × C5 × C1
= 777
13.
If the digits are not allowed to repeat in any number formed by using the digits 0, 2, 4, 6, 8, then the number
of all numbers greater than 10,000 is equal to _______.
[JEE MAINS - ONLINE - 2021]
Official Ans. by NTA (96)
Sol.
2,4,6,8
4
4 3 2 1
= 4 × 4 × 3 × 2 = 96
14.
There are 5 students in class 10, 6 students in class 11 and 8 students in class 12. If the number of ways, in
which 10 students can be selected from them so as to include at least 2 students from each class and at most
5 students from the total 11 students of class 10 and 11 is 100 k, then k is equal to _______.
[JEE MAINS - ONLINE - 2021]
Number of selection
Þ
Þ
th
th
th
10
5
2
11
6
3
12
8
5
6
8
5 Þ C2 × C3 × C5
2
2
6 Þ C2 × C2 × C6
3
2
5 Þ C3 × C2 × C5
5
6
8
5
6
8
Total number of ways = 23800
According to question
100 K = 23800
K = 238
10
15.
11
Let n be a non-negative integer. Then the number of divisors of the form “4n + 1” of the number (10) . (11) .
13
(13) is equal to_____ .
[JEE MAINS - ONLINE - 2021]
Official Ans. by NTA (924)
10
10
11
13
Sol. N = 2 × 5 × 11 × 13
Now, power of 2 must be zero,
power of 5 can be anything,
28
JPR\COMP.251\Allen(IIT-JEE Wing)\2020–21\Nurture\Mathematics\UNIT - 4
Official Ans. by NTA (238)
Sol. Class :
Total student
Permutation - Combination
power of 13 can be anything.
But, power of 11 should be even.
So, required number of divisors is
1 × 11 × 14 × 6 = 924
16.
15
If 1 P1 ∗ 2·2 P2 ∗ 3·3 P3 ∗ ... + 15 ·
q
P15 = Pr – s, 0 £ s £ 1, then
q+s
Cr–s is equal to ________.
[JEE MAINS - ONLINE - 2021]
Official Ans. by NTA (136)
Sol.
1
P1 ∗ 2·2 P2 ∗ 3·3 P3 ∗ ... + 15 ·
15
P15
= 1! + 2.2! + 3.3! + ….15 × 15!
15
=
å ∋r ∗ 1 , 1( r !
r <1
15
=
Þ
å ∋r ∗ 1(!, ∋r (!
r <1
= 16! – 1
16
= P16 – 1
q = r = 16, s = 1
q+s
17
Cr–s = C15 = 136
17.
The number of three-digit even numbers, formed by the digits 0, 1, 3, 4, 6, 7 if the repetition of digits is not
allowed, is ________.
[JEE MAINS - ONLINE - 2021]
Official Ans. by NTA (52)
Sol. (i) When ‘0’ is at unit place
­
5
­
4
0
Number of numbers = 20
(ii) When 4 or 6 are at unit place
OX
4 × 4
4,6
­
2
Number of numbers = 32
So number of numbers = 52
18.
JPR\COMP.251\Allen(IIT-JEE Wing)\2020–21\Nurture\Mathematics\UNIT - 4
The sum of all 3-digit numbers less than or equal to 500, that are formed without using the digit “1” and they
all are multiple of 11, is _______.
[JEE MAINS - ONLINE - 2021]
Official Ans. by NTA (7744)
So1. 209, 220, 231, ..., 495
Sum =
27
∋ 209 ∗ 495( < 9504
2
Number containing 1 at unit place
th
Number containing 1 at 10 place
2 3 1
3 4 1
4 5 1
3 1 9
4 1 8
Required = 9501 – (231 + 341 + 451 + 319 + 418)
29
JEE-Mathematics
7744
19.
A number is called a palindrome if it reads the same backward as well as forward. For example 285582 is a six
digit palindrome. The number of six digit palindromes, which are divisible by 55, is _________.
[JEE MAINS - ONLINE - 2021]
Official Ans. by NTA (100)
5 a b
Sol.
b a 5
It is always divisible by 5 and 11.
So, required number = 10 × 10 = 100
20.
The number of six letter words (with or without meaning), formed using all the letters of the word ‘VOWELS’,
so that all the consonants never come together, is _________.
[JEE MAINS - ONLINE - 2021]
Official Ans. by NTA (576)
Sol. VOWELS
2 Vowels
4 Consonants
JPR\COMP.251\Allen(IIT-JEE Wing)\2020–21\Nurture\Mathematics\UNIT - 4
All Consonants should not be together
= Total – All consonants together,
= 6! – 3! 4! = 576
30
Permutation - Combination
SECTION - 1 : SINGLE CHOICE CORRECT QUESTIONS
1.
In a shop there are five types of ice–creams available. A child buys six ice–creams.
[AIEEE 2008]
10
Statement –1 – The number of different ways the child can buy the six ice–creams is C5.
Statement –2 – The number of different ways the child can buy the six ice–creams is equal to the number
of different ways of arranging 6 A’s and 4 B’s in a row.
(1) Statement –1 is false, Statement –2 is true
(2) Statement–1 is true, Statement–2 is true; Statement–2 is a correct explanation for Statement–1
(3) Statement–1 is true, Statement–2 is true; Statement–2 is not a correct explanation for Statement–1
(4) Statement–1 is true, Statement–2 is false
Ans. (1)
10
Sol. Boy can buy icecream in C4 number of ways.
2.
There are two urns. Urn A has 3 distinct red balls and urn B has 9 distinct blue balls. From each urn two balls
are taken out at random and then transferred to the other. The number of ways in which this can be done is
:[AIEEE-2010]
(1) 3
(2) 36
(3) 66
(4) 108
Ans. (4)
Sol.
UrnA
UrnB
3 Red
Balls
9 Blue
Balls
3
9
Number of ways C2 × C2 = 3 × 36 = 108
Option (4) is correct
3.
If the number of 5-element subsets of the set A = {a1, a2, ........, a20} of 20 distinct elements is k times the
number of 5 - element subsets containing a4 then k is :
[AIEEE ONLINE-2012]
(1) 5
(2)
Ans. (3)
20
Sol.
C5 = K ×
19
20
7
(3) 4
(4)
10
3
C4
20!
19!
= k×
15! 5!
15! 4!
JPR\COMP.251\Allen(IIT-JEE Wing)\2020–21\Nurture\Mathematics\UNIT - 4
k=
20
=4
5
4.
Assuming the balls to be identical except for difference in colours, the number of ways in which one or more
balls can be selected from 10 white, 9 green and 7 black balls is :[AIEEE-2012]
(1) 879
(2) 880
(3) 629
(4) 630
Ans. (1)
Sol. Number of ways in which one or more balls can be selected
= 11 × 10 × 8 – 1
= 879
Option (1) is correct
31
JEE-Mathematics
On the sides AB, BC, CA of a DABC 3, 4, 5 distinct points (excluding vertices A, B, C) are respectively chosen.
The number of triangles that can be constructed using these chosen points as vertices are :
[JEE-MAIN ONLINE-2013]
(1) 210
(2) 215
(3) 220
(4) 205
Ans. (4)
12
5
4
3
Sol.
C3 – C3 – C3 – C3
220 – 10 – 4 – 1 = 205
5.
C
A
B
6.
8-digit numbers are formed using the digits 1, 1, 2, 2, 2, 3, 4, 4. The number of such number in which the
odd digits do not occupy odd places, is :
[JEE-MAIN ONLINE-2014]
(1) 60
(2) 48
(3) 160
(4) 120
Ans. (4)
4
Sol. Odd digits 1, 1, 3 occupy four even places so number of arrangements =
P3
= 12
2!
Remaining 2, 2, 2, 4, 4 can be arranged at 5 places
5!
= 3! 2! = 10 No. of ways
Total = 12 × 10 = 120
7.
If the four letter words (need not be meaningful) are to be formed using the letters from the word
‘‘MEDITERRANEAN’’ such that the first letter is R and the fourth letter is E, then the total number of all such
words is :
[JEE-MAIN ONLINE-2016]
Ans. (4)
Sol. M E D I T R A N
1 3111222
R––– E
Case-1 Both diff.
8
P2 = 56
Case-2 Both identical
3
C1 = 3
Total = 56 + 3
= 59
32
(2)
11!
(2!)3
(3) 110
(4) 59
JPR\COMP.251\Allen(IIT-JEE Wing)\2020–21\Nurture\Mathematics\UNIT - 4
(1) 56!
Permutation - Combination
8.
A man X has 7 friends, 4 of them are ladies and 3 are men. His wife Y also has 7 friends, 3 of them are ladies
and 4 are men. Assume X and Y have no common friends. Then the total number of ways in which X and Y
together can throw a party inviting 3 ladies and 3 men, so that 3 friends of each of X and Y are in this party is
[JEE (Main)-2017]
(1) 469
(2) 484
(3) 485
(4) 468
Ans. (3)
x
7
y
7
Sol.
4L
3
3M
3L
3
4M
( 4 C3 3 C 0 + 4 C 2 3 C1 + 4 C1 3 C 2 + 4 C0 3C 3 )
Number of ways =
2
2
2
2
2
2
= 4 + 6 . 3 + 4 .3 + 1
= 16 + 324 + 144 + 1 = 485
9.
Let S be the set of all triangles in the xy-plane, each having one vertex at the origin and the other two vertices
lie on coordinate axes with integral coordinates. If each triangle in S has area 50sq. units, then the number
of elements in the set S is :
[JEE-MAIN ONLINE-2019]
(1) 9
(2) 18
(3) 32
(4) 36
Ans. (4)
Sol. Let A(a,0) and B(0,b)
be the vectors of the given triangle AOB
Þ
|ab| = 100
Þ
Number of triangles
= 4 × (number of divisors of 100)
= 4 × 9 = 36
10.
Consider three boxes, each containing 10 balls labelled 1,2,....,10. Suppose one ball is randomly drawn from
th
each of the boxes. Denote by ni, the label of the ball drawn from the i box, (i = 1, 2, 3). Then, the number
of ways in which the balls can be chosen such that n1 < n2 < n3 is :
[JEE-MAIN ONLINE-2019]
(1) 82
(2) 240
(3) 164
(4) 120
Ans. (4)
Sol. No. of ways = 10C3 = 120
11.
The number of four-digit numbers strictly greater than 4321 that can be formed using the digits 0, 1, 2, 3,
4, 5 (repetition of digits is allowed) is :
[JEE-MAIN ONLINE2019]
(1) 288
(2) 306
(3) 360
(4*) 310
Ans. (4)
Sol. 0, 1, 2, 3, 4, 5
JPR\COMP.251\Allen(IIT-JEE Wing)\2020–21\Nurture\Mathematics\UNIT - 4
4
4 or 5
2
4
6
3 3, 4, 5
Total ways = 72
6
= 3 × 6 = 18
3
4
3
2
5
=4
3
= 6 = 216
Total = 310
33
JEE-Mathematics
12.
Some identical balls are arranged in rows to form an equilateral triangle. The first row consists of one ball,
the second row consists of two balls and so on. If 99 more identical balls are addded to the total number
of balls used in forming the equilaterial triangle, then all these balls can be arranged in a square whose each
side contains exactly 2 balls less than the number of balls each side of the triangle contains. Then the number
of balls used to form the equilateral triangle is :
[JEE-MAIN ONLINE2019]
(1*) 190
(2) 262
(3) 225
(4) 157
Ans. (1)
Sol. Number of balls used in the triangle
=
\
\
n(n + 1)
2
n(n + 1)
+ 99 = (n - 2)2
2
2
n – 9n = 190
n(n – 9) = 190 Þ n = 19
n(n + 1) 19 ´ 20
=
= 190
2
2
13.
The number of 6 digit numbers that can be formed using the digits 0, 1, 2, 5, 7 and 9 which are divisible by 11
and no digit is repeated, is :
[JEE-MAIN ONLINE2019]
(1) 36
(2*) 60
(3) 48
(4) 72
Ans. (2)
Sol. Let x1, x2, x3, x4 be the 6 digit letter then
(x1 + x3 + x5) – (x2 + x4 + x6) should be multiple of 11
only possibility is (9, 2, 1) and (7, 5, 0)
Which will give 3 3 + 2 2 3 = 36 + 24 = 60
14.
34
=
1 21
é C0 +
2ë
=
1 21
.2 = 220.
2
21
C1 + .... +
21
C10 +
21
C11 + .... +
21
C 21 ùû
JPR\COMP.251\Allen(IIT-JEE Wing)\2020–21\Nurture\Mathematics\UNIT - 4
The number of ways of choosing 10 objects out of 31 objects of which 10 are identical and the remaining
21 are distinct, is :
[JEE-MAIN ONLINE2019]
20
20
20
21
(1*) 2
(2) 2 – 1
(3) 2 + 1
(4) 2
Ans. (1)
Sol. 21 distinct + 10 identical objects
Case : (1) 10 identical chosen = 1 case
21
Case : (2) 9 identical chosen +1 distanct = 1. C1 case
21
Case : (3) 8 identical chosen +2 distanct = 1 . C2 and so on
21
21
21
Total cases = C0 + C1 ....... + C10
Permutation - Combination
15.
Two families with three members each and one family with four members are to be seated in a row. In how
many ways can they be seated so that the same family members are not separated ?
[JEE MAINS - ONLINE - 2020]
(1) 2!3!4!
(2)
(3!)3.(4!)
(3)
(3!)2.(4!)
(4) 3!(4!)3
Official Ans. by NTA (2)
Sol. Total numbers in three familes = 3 + 3 + 4 = 10
so total arrangement = 10!
Family1 Family 2 Family 3
Favourable cases
3
3
4
=
\
3!
Arrangment of 3 Families
3! ´ 3! ´ 4 !
Interval Arrangment of families members
Probability of same family memebers are together =
3! 3! 3! 4 !
1
=
10!
700
so option(2) is correct.
16.
x y z
A natural number has prime factorization given by n = 2 3 5 , where y and z are such that y + z = 5 and
5
, y > z. Then the number of odd divisors of n, including 1, is :[JEE MAINS - ONLINE - 2021]
6
(1) 11
(2) 6
(3) 6x
(4) 12
Official Ans. by NTA (4)
Sol. Ans. (4)
Sol. y + z = 5
–1
–1
y +z =
1 1 5
+ =
y z 6
Þ
Þ
JPR\COMP.251\Allen(IIT-JEE Wing)\2020–21\Nurture\Mathematics\UNIT - 4
17.
y>z
y = 3, z = 2
x 3 2
n = 2 .3 .5 = (2.2.2 ...) (3.3.3) (5.5)
Number of odd divisors = 4 × 3 = 12
Consider a rectangle ABCD having 5, 7, 6, 9 points in the interior of the line segments AB, CD, BC, DA respectively.
Let a be the number of triangles having these points from different sides as vertices and b be the number of
quadrilaterals having these points from different sides as vertices. Then (b – a) is equal to :
(1) 795
(2) 1173
Official Ans by NTA (4)
Sol. a = Number of triangles
a = 5·6·7 + 5·7·9 + 5·6·9 + 6·7·9
= 210 + 315 + 270 + 378
= 1173
b = Number of Quadrilateral
b = 5·6·7·9 = 1890
b – a = 1890 – 1173 = 717
(3) 1890
A
[JEE MAINS - ONLINE - 2021]
(4) 717
5 Pts
6 Pts
9 Pts
D
B
7 Pts
C
35
JEE-Mathematics
18.
Let P1, P2, ….., P15 be 15 points on a circle. The number of distinct triangles formed by points P i, Pj, Pk such
that i + j + k ¹ 15, is :
[JEE MAINS - ONLINE - 2021]
(1) 12
(2) 419
(3) 443
(4) 455
Official Ans. by NTA (3)
15
Sol. Total Number of Triangles =
C3
i + j + k = 15 (Given)
5 Cases
i j k
1 2 12
1 3 11
1 4 10
1 5 9
1 6
4 Cases
3 Cases
i
i
j
k
2 3 10
2 4
2 5
9
8
2 6
7
j
k
3 4 8
3 5 7
1 Cases
i j k
4 5 6
8
15
Number of Possible triangles using the vertices Pi, Pj, Pk such that i + j + k ¹ 15 is equal to C3 – 12 = 443
Option (3)
SECTION - 2 : NUMERICAL ANSWER BASED QUESTIONS
1.
The number of 4 letter words (with or without meaning) that can be formed from the eleven letters of the word
'EXAMINATION' is _______.
[JEE-MAIN ONLINE2020]
NTA Ans. (2454)
ALLEN Ans. (2454.00)
Sol. N ® 2, A ® 2, I ® 2, E, X, M, T, O ® 1
Category
2 alike of one kind
Selection
3
and 2 alike of other kind
2 alike and 2 different
All 4 different
3
C2 = 3
C1 ´7 C 2
8
Arrangement
4!
= 18
2! 2!
4!
3
= 756
C1 ´7 C 2 ´
2!
C4
3´
8
C4 ´ 4 ! = 1680
Total = 2454
Ans. 2454.00
Let S = {1, 2, 3, 4, 5, 6, 9}. Then the number of elements in the set T = {A Í S : A ¹ f and the sum of all
the elements of A is not a multiple of 3} is _________.
[JEE MAINS - ONLINE - 2021]
Official Ans. by NTA (80)
Sol.
3n type
® 3, 6, 9 = P
3n – 1 type ® 2, 5 = Q
3n – 2 type ® 1,4 = R
number of subset of S containing one element
2
2
which are not divisible by 3 = C1 + C1 = 4
number of subset of S containing two numbers
whose some is not divisible by 3
= 3 C1 ≥ 2C1 ∗ 3 C1 ≥ 2C1 ∗ 2C 2 ∗ 2C 2 < 14
36
JPR\COMP.251\Allen(IIT-JEE Wing)\2020–21\Nurture\Mathematics\UNIT - 4
2.
Permutation - Combination
number of subsets containing 3 elements whose
sum is not divisible by 3
< 3 C 2 ≥ 4 C1 ∗ ∋ 2 C 2 ≥ 2C1 ( 2 ∗ 3 C1 ∋ 2 C 2 ∗ 2 C 2 ( < 22
number of subsets containing 4 elements whose
sum is not divisible by 3
< 3 C 3 ≥ 4 C1 ∗ 3 C 2 ∋ 2 C 2 ∗ 2C 2 ( ∗ ∋ 3 C12C1 ≥ 2 C 2 ( 2
= 4 + 6 + 12 = 22.
number of subsets of S containing 5 elements
whose sum is not divisible by 3.
= 3 C 3 ∋ 2 C 2 ∗ 2 C 2 ( ∗ ∋ 3 C 22C1 ≥ 2 C 2 (≥ 2 = 2 + 12 = 14
number of subsets of S containing 6 elements
whose sum is not divisible by 3 = 4
Þ Total subsets of Set A whose sum of digits is
not divisible by 3 = 4 + 14 + 22 + 22 + 14 + 4 = 80.
3.
All the arrangements, with or without meaning, of the word FARMER are written excluding any word that has
two R appearing together. The arrangements are listed serially in the alphabetic order as in the English dictionary.
Then the serial number of the word FARMER in this list is ___.
[JEE MAINS - ONLINE - 2021]
Official Ans. by NTA (77)
Sol. FARMER (6)
A, E, F, M, R, R
A
E
F
F
F
F
A
A
A
A
E
M
R
R
E
M
E
R
5
, 4 < 60 , 24 < 36
2
3
, 2 < 3, 2 < 1
2
=1
=2
=1
JPR\COMP.251\Allen(IIT-JEE Wing)\2020–21\Nurture\Mathematics\UNIT - 4
77
37
JPR\COMP.251\Allen(IIT-JEE Wing)\2020–21\Nurture\Mathematics\UNIT - 4
JEE-Mathematics
38
Permutation - Combination
1.
Consider all possible permutations of the letters of the word ENDEANOEL.
[JEE 2008]
Match the Statements / Expressions in Column I with the Statements / Expressions in Column II.
(A)
(B)
(C)
(D)
Column I
Column II
The number of permutations containing the word ENDEA is
The number of permutations in which the letter E occurs in the first
and the last positions is
The number of permutations in which none of the letters D, L, N
occurs in the last five positions is
The number of permutations in which the letters A, E, O occur only in
odd positions is
(p)
(q)
5!
2 × 5!
(r)
7 × 5!
(s)
21 × 5!
Ans. (A)®(p), (B)®(s), (C)®(q), (D)®(q)
Sol. A. If ENDEA is fixed word, then assume this as a single letter. Total number of letters = 5
Total number of arrangements = 5!
B. If E is at first and last places, then total number of permutations =
7!
= 21 × 5!
2!
C. If D, L, N are not in last five positions
¬ D, L, N, N ® ¬ E, E, A, O ®
4 ! 5!
´ = 2 ´ 5!
2! 3!
D. Total number of odd positions = 5
Total number of permutations =
5!
3!
Total number of even positions = 4
Permutations of AEEEO are
\ Number of permutation of N, N, D, L =
Þ Total number of permutations =
4!
2!
5! 4 !
´ = 2 ´ 5!
3! 2!
2.
The number of seven digit integers, with sum of the digits equal to 10 and formed by using the digits 1, 2 and 3
only is [JEE 2009]
(A) 55
(B) 66
(C) 77
(D) 88
Ans. (C)
Sol. There are two possible cases
Case I Five 1’s, one 2’s one 3’s
7!
= 42
5!
Case II Four 1’s, three 2’s
JPR\COMP.251\Allen(IIT-JEE Wing)\2020–21\Nurture\Mathematics\UNIT - 4
Number of numbers =
7!
= 35
4!3!
\ Total number of numbers = 42 + 35 = 77.
Number of numbers =
39
JEE-Mathematics
3.
Let S = {1, 2, 3, 4}. The total number of unordered pairs of disjoint subsets of S is equal to - [JEE 2010]
(A) 25
(B) 34
(C) 42
(D) 41
Ans. (D)
Sol. Case- I : The number of elements in the pairs can be 1,1; 1,2; 1,3,; 2,2
C2 . 2 C2
= 25
2
4
Case- II : Number of pairs with f as one of subsets = 2 = 16
\ Total pairs = 25 + 16 = 41
4
4
3
4
3
4
= C2 + C1 × C2 + C1 × C3 +
4.
The total number of ways in which 5 balls of different colours can be distributed among 3 persons so that
each person gets at least one ball is [JEE 2012]
(A) 75
(B) 150
(C) 210
(D) 243
Ans. (B)
Sol. Objects
Groups
Objects
Groups
Distinct
Distinct
Identical
Identical
Distinct
Identical
Identical
Distinct
Description of Situation : Here, 5 distinct balls are distributed amongst 3 persons so that each gets at least
on ball. i.e. Distinct ® Distinct
So, we should make cases.
Case I
A B Cü
A B Cü
ý Case II
ý
1 1 2þ
1 2 2þ
Number of ways to distribute 5 balls
3! ö æ 5
3! ö
æ5
4
3
4
2
= ç C1 . C1 . C 3 ´ ÷ + ç C1 . C 2 . C 2 ´ ÷
2! ø è
2! ø
è
= 60 + 90 = 150
Paragraph for Question 5 and 6
Let an denotes the number of all n-digit positive integers formed by the digits 0, 1 or both such that no consecutive
digits in them are 0. Let bn = the number of such n-digit integers ending with digit 1 and c n = the number of such
n-digit integers ending with digit 0.
The value of b6 is
(A) 7
Ans. (B)
[JEE 2012]
(B) 8
(C) 9
(D) 11
(C) b17 ¹ b16 + c16
[JEE 2012]
(D) a17 = c17 + b16
*6.
Which of the following is correct ?
(A) a17 = a16 + a15
(B) c17 ¹ c16 + c15
Ans. (A)
Solution for Q.5 & Q.6
For an
The first digit should be 1
For bn
1_ _ _ _ _ .... _ _1
1442443
( n -2 Places )
Last digit is 1. so bn is equal to number of ways of an–1 (i.e. remaining (n – 1) places)
bn = an–1
For cn
40
JPR\COMP.251\Allen(IIT-JEE Wing)\2020–21\Nurture\Mathematics\UNIT - 4
*5.
Permutation - Combination
Last digit is 0 so second last digit must be 1
So cn = an–2
bn + c n = a n
So an = an–1 + an–2
Similarly bn = bn–1 + bn–2
(5) Ans.(B)
a1 = 1, a2 = 2
So a3 = 3,
a4 = 5 a5 = 8
Þ b6 = a 5 = 8
(6) Ans.(A)
an = an–1 + an–2
put n = 17
a17 = a16 + a15
(A) is correct
cn = cn–1 + cn–2
So put n = 17
c17 = c16 + c15
(B) is incorrect
bn = bn–1 + bn–2
put n = 17
b17 = b16 + b15
(C) is incorrect
a17 = a16 + a15
while (D) says a17 = a15 + a15(D) is incorrect
*7.
Let n1 < n2 < n3 < n4 < n5 be positive integers such that n1 + n2 + n3 + n4 + n5 = 20. Then the number of
such distinct arrangements (n1, n2, n3, n4, n5) is
[JEE 2014]
Ans. 7
Sol. As, n1 ³ 1, n2 ³ 2, n3 ³ 3, n4 ³ 4, n5 ³ 5
Let n1 – 1 = x1 ³ 0, n2 – 2 = x2 ³ 0, .... , n5 – 5 = x5 ³ 0
Þ
New equation will be
x1 + 1 + x2 + 2 + ... + x5 + 5 = 20
Þ
x1 + x2 + x3 + x4 + x5 = 20 – 15 = 5
Now, x1 £ x2 £ x3 £ x4 £ x5
x1
0
x2
0
x3
0
x4
0
x5
5
0
0
0
1
4
0
0
0
0
0
1
2
1
3
3
0
0
0
1
1
1
2
1
2
2
1
1
1
1
1
So, 7 possible cases will be there.
Let n ³ 2 be an integer. Take n distinct points on a circle and join each pair of points by a line segment. Colour
the line segment joining every pair of adjacent points by blue and the rest by red. If the number of red and blue
line segments are equal, then the value of n is :
[JEE 2014]
Ans. 5
n
Sol. Number of blue line segments = C2 – n
n
\
C2 – n = n
JPR\COMP.251\Allen(IIT-JEE Wing)\2020–21\Nurture\Mathematics\UNIT - 4
8.
Þ
Þ
n(n - 1)
= 2n
2
n=5
41
JEE-Mathematics
9.
Six cards and six envelopes are numbered 1, 2, 3, 4, 5, 6 and cards are to be placed in envelopes so that each
envelope contains exactly one card and no card is placed in the envelope bearing the same number and
moreover the card numbered 1 is always placed in envelope numbered 2. Then the number of ways it can be
done is
[JEE 2014]
(A) 264
(B) 265
(C) 53
(D) 67
Ans. (C)
Sol.
D6
= 53,
5
where D6 = Derangement of 6 persons
10.
Let n be the number of ways in which 5 boys and 5 girls can stand in a queue in such a way that all the girls
stand consecutively in the queue. Let m be the number of ways in which 5 boys and 5 girls can stand in a queue
m
in such a way that exactly four girls stand consecutively in the queue. Then the value of
is
n
[JEE 2015]
Ans. 5
Sol. Here, __ B1 __ B2 __ B3 __ B4 __ B5 __
Out of 5 girls, 4 girls are together and 1 girl is separate. Now, to select 2 positions out of 6 positions between boys
6
= C2
.....(i)
5
4 girls are to be selected out 5 = C4
....(ii)
Now, 2 groups of girls can be arranged in 2! ways.
...(iii)
Also, the group of 4 girls and 5 boys in arranged in 4! × 5! ways.
...(iv)
6
5
Now, total number of ways C2 × C4 × 2! × 4! × 5! [From Questions, (i), (ii), (iii) and (iv)]
6
5
\
m = C2 × C4 × 2! × 4! × 5!
and n = 5! × 6!
m 6 C 2 ´ 6 C 4 ´ 2!´ 5! 15 ´ 5 ´ 2 ´ 4 !
=
=
=5
n
6!´ 5!
6 ´ 5 ´ 4!
Þ
11.
A debate club consists of 6 girls and 4 boys. A team of 4 members is to be selected from this club including the
selection of a captain (from among these 4 members) for the team. If the team has to include at most one boy,
then the number of ways of selecting the team is
[JEE 2016]
(A) 380
(B) 320
(C) 260
(D) 95
Ans. (A)
Sol. We have, 6 girls and 4 boys. To select 4 members (atmost one boy)
6
4
6
i.e. (1 boy and 3 girls) or (4 girls) = C3. C1 + C4
....(i)
4
Now, selection of captain, from these 4 members = C1
...(ii)
\ Number of ways to select 4 members (including the selecting of a captain, from these 4 members)
6
4
6
4
= ( C3. C1 + C4) C1
= (20 × 4 + 15) × 4 = 380
12.
no other letter is repeated. Then
y
=
9x
Ans. (5)
Sol. A, B, C, D, E, F, G, H, I, J
k = 10!
y=
10
C 2 ´ 2 C1 ´
10! 10 ´ 9
=
´ 10!
2!
2
y 10 ´ 9 ´ 10!
=
=5
9x 2 ´ 10! ´ 9
42
[JEE 2017]
JPR\COMP.251\Allen(IIT-JEE Wing)\2020–21\Nurture\Mathematics\UNIT - 4
Words of length are formed using the letters A, B, C, D, E, F, G, H, I, J. Let x be the number of such words
where no letter is repeated; and let y be the number of such words where exactly one letter is repeated twice and
Permutation - Combination
13.
Let S = {1, 2, 3, ..... 9}. For k = 1, 2, ...., 5, let N k be the number of subsets of S, each containing five elements
out of which exactly k are odd. Then N 1 + N2 + N3 + N4 + N5 =
[JEE 2017]
(A) 125
(B) 252
(C) 210
(D) 126
Ans. (D)
4
5
5
4
5
4
5
4
5
Sol. C4 C1 + C2 C3 + C3 C2 + C4 C1 + C5
9
= C5 = 126
14.
The number of 5 digit numbers which are divisible by 4, with digits from the set {1,2,3,4,5} and the repetition
of digits is allowed, is _____ .
[JEE 2018]
Ans. (625)
Sol. Last 2 digits must be divisible by 4
– – – 12 = 125
– – – 24 = 125
– – – 32 = 125
– – – 52 = 125
– – – 44 = 125
= 625
15.
In a high school, a committee has to be formed from a group of 6 boys M 1, M2, M3, M4, M5, M6 and 5 girls G1,
G2, G3, G4, G5.
[JEE 2018]
(i) Let a1 be the total number of ways in which the committee can be formed such that the committee has 5
members, having exactly 3 boys and 2 girls.
(ii) Let a2 be the total number of ways in which the committee can be formed such that the committee has
at least 2 members and having an equal number of boys and girls.
(iii) Let a 3 be the total number of ways in which the committee can be formed such that the committee has 5
members, at least 2 of them being girls.
(iv) Let a4 be the total number of ways in which the committee can be formed such that the committee has 4
members, having at least 2 girls and such that M 1 and G1 are NOT in the commitee together.
LIST-I
(P)
(Q)
(R)
(S)
The value of a1 is
The value of a2 is
The value of a3 is
The value of a4 is
LIST-II
1.
2.
3.
4.
5.
6.
136
189
192
200
381
461
JPR\COMP.251\Allen(IIT-JEE Wing)\2020–21\Nurture\Mathematics\UNIT - 4
The correct option is :
(A) P ® 4; Q ® 6; R ® 2; S ® 1
(B) P ® 1; Q ® 4; R ® 2; S ® 3
(C) P ® 4; Q ® 6; R ® 5; S ® 2
(D) P ® 4; Q ® 2; R ® 3; S ® 1
Ans. (C)
43
JEE-Mathematics
16.
Five person A,B,C,D and E are seated in a ciruclar arrangement. If each of them is given a hat of one of the
three colours red, blue and green ,then the number of ways of distributing the hats such that the persons seated
in adjacent seats get different coloured hats is
[JEE-Advanced 2019]
Ans. (30.00)
A
Sol. When 1R, 2B, 2G
B
E
5C1 × 2 = 10
Other possibilities
1B, 2R, 2G
D
C
or
1G, 2R, 2B
So total no. of ways = 3 × 10 = 30
2020, Paper-2
17. An engineer is required to visit a factory for exactly four days during the first 15 days of every month and it is
mandatory that no two visits take place on consecutive days. Then the number of all possible ways in which
such visits to the factory can be made by the engineer during 1-15 June 2021 is _____
[JEE-Advanced 2020]
Ans. 495.00
Sol. Selection of 4 days out of 15 days such that no two of them are consecutive
=
=
15–4+1
C4 =
12
C4
12 ´ 11 ´ 10 ´ 9
= 11 ´ 5 ´ 9 = 495
4´3´2
18.
In a hotel, four rooms are available. Six persons are to be accommodated in these four rooms in such a way
that each of these rooms contains at least one person and at most two persons. Then the number of all possible
ways in which this can be done is _____
[JEE-Advanced 2020]
Ans. 1080.00
6!
Sol. required ways = 2! 2! 1! 1! 2! 2! ´ 4! = 1080
2021, Paper-2
19.
Let
and
S1 = {(i, j, k) : i, j, k Î {1, 2, ...., 10}},
S2 = {(i, j) : 1 £ i < j + 2 £ 10, i, j Î {1, 2, ....., 10}},
S3 = {(i, j, k, l) : 1 £ i < j < k < l, i, j, k, l Î {1, 2, ...., 10}}
S4 = {(i, j, k, l) : i, j, k and l are distinct elements in {1, 2, ......, 10}}.
If the total number of elements in the set Sr is nr, r = 1, 2, 3, 4 then which of the following statements is(are)
TRUE ?
(B) n2 = 44
Ans. (ABD)
Sol. n1 = 10 × 10 × 10 = 1000
ìi <1↑ 1′ j ′ 8 i < 2 ↑ i ′ j ′ 8 ï
ü
ï
ï
ï
ï
ï
ï
ï
i
<
3
↑
2
′
j
′
8
i
<
4
↑
3
′
j
′
8
ï
ï
ï
ï
ï
n2 ↑ ï
íi < 5 ↑ 4 ′ j ′ 8 i < 6 ↑ 5 ′ j ′ 8 ý
ï
ï
ï
ï
ï
i < 7 ↑ 6 ′ j ′ 8 i < 8 ↑ 7 ′ j ′ 8ï
ï
ï
ï
ï
ï
ï
i
<
9
↑
j
<
8
ï
ï
î
þ
44
(C) n3 = 220
n4
< 420
12
[JEE-Advanced 2021]
(D)
JPR\COMP.251\Allen(IIT-JEE Wing)\2020–21\Nurture\Mathematics\UNIT - 4
(A) n1 = 1000
Permutation - Combination
(1 + 2 + ...... + 7 + 8) + 8 = 44
n3 < 10 C 4 <
10 × 8 × 8 × 7
< 210
1× 2 × 3 × 4
JPR\COMP.251\Allen(IIT-JEE Wing)\2020–21\Nurture\Mathematics\UNIT - 4
n4 10 C 4 × 4 10
<
< c 4 × 2 < 420
12
12
45