Supplement homework
1.
Evaluate each of the following.
(a) 7! = 5040
(b) 8P2 = 56
12!
(d) P(14, 5) = 240240
(e)
= 924
(c) 4! x 3! = 144
3!6!5!
2. Simplify:
(a)
𝑟!
(𝑟 −1)
!
=r
(b)
(𝑛−𝑟)!
(𝑛−𝑟+1)!
3. Solve for n, if P(n,r) = 5 P(n–1,r–1). N =5
4. How many distinct arrangements of the letters in PERMUTATIONS can you make?
(a) taking all the letters at a time 12!/2!
(b) taking seven letters at a time (12P7)/2!
(c) taking three letters at a time (12P3)/2!
5.
How many distinct four-digit odd numbers can be formed from the digits in the number 6 738
195? 5(6P3)
6.
Manpreet has 6 romance novels, 4 fiction novels, and 9 war novels on a shelf. In how many ways
can she arrange her novels on the shelf if novels of the same genre are to be kept together?
(6!x4!x9!)(3!)
7.
Ten beads of different colours are strung on a string to make a necklace. How many different
arrangements are there?
10!/2
1. A bookstore has 1000 hardcover books and 4500 paperback books for sale.
a. How many books does the store have for sale? 5500
b. If Stan plans to buy 1 hardcover and 1 paperback, then how many different choices can he make?
1000x4500 = 4500000
2. A student has 3 homework assignments to do. In how many different orders can they be completed?
3! = 6
3. At one time, postal codes consisted of 2 letters followed by 5 digits. How many different postal
codes were possible? (Assume that different letter combinations give rise to different postal codes.)
26x25x10x10x10x10x10
4. In a small European county, the licenses on cars consist of 4 letters. Suppose the government wants
to change the licenses to numbers only. How many digits should the license plates be to have the same
number of plates?
26x26x26x26=456976 at least 6 number
5. An election for an executive committee of the student council has 2 candidates for president, 4
candidates for vice president, 3 candidates for treasurer, and 2 candidates for secretary. How many
different executive committees are possible?
2x4x3x2 =48
6. A tennis pro has 3 pairs of shoes, 6 pairs of socks, 5 pairs of shorts and 8 shirts. How many different
outfits can she wear?
3x6x5x8=720
7. How many different 3‐digit positive integers are there in which the tens digit and the ones digit are
different?
9x10x9 = 810
8. How many 3‐digit positive integers are there in which:
a. All the digits are odd? 9x10x5 = 450
b. All the digits are even? 9x10x5 = 450
9. a. How many 2‐digit numbers are there that are not divisible by either 2 or 5?
End with 1,3,7,9 9x4 = 36
b How many odd positive integers less than 1000 are there with all digits distinct?
Middle is 0 and middle is not 0: 8x1x5 + 7x8x5 = 320
10. The math department is taking its photos for the yearbook. If there are 12 teachers in a row (one
person per seat):
a. How many arrangements are possible? 12!
b. How many arrangements are possible if Mr. Fabroa must be on the right end of the row and
Mr. Scuili and Ms. Cea want to sit beside each other? 11 left, count S & C as one: P(10,10)x2
c. How many arrangements are possible if Ms. Cea and Ms. Do do not sit beside each other?
All – Cea & Do together = 12! – 2P(11,11) = 399168000
11. How many permutations of the digits {1, 2, 3, 4, 5, 6}, taken four at a time:
a. Do not contain 6?
P(5,4)
b. Has the 1 in the first position? 1x5x4x3= 60
c. Has the 1 in the third position?
5x4x1x3 = 60
d. Have the 3 in the second position and the 5 in the third position? 4x1x1x3 = 12
e. Are odd?
5x4x3x3 = 180
f. Are divisible by 2 or 5?
5x4x3x2= 120
12. How many permutations of the numbers {1, 2, 3, 4, 5, 6}:
a. Begin with an even number? 3x5x4x3x2x1 = 3x5! = 360
b. Begin with an even number and end with an odd number? 3x4!x3 = 216
c. Begin with an odd number and end with an odd number? 3x4!x2 = 144
13. How many permutations are there of the letters in the word ISOSCELES if: 9 letters 3S 2E
a. The C occurs in the middle? 8! ÷ (3! x 2!)
b. Begins with the O but does not end with the L?
1 choice for the first, 4 for last = 1x 7! ÷ (3! X 2!) x 4
c. The E’s are adjacent? Count E as one, 8!
d. The O occurs in the 3rd, 4th, or 7th position, the C occurs in the 3rd, 4th, or 7 th position, and
the L occurs in the 3rd, 4th, or 7th position? 3! For O, C, L and 6 left = 3! x 6! ÷ (3! 2!)
14. How many permutations of the letters in the name OLIVER has the vowels falling in alphabetical
order? (but not necessarily adjacent) EIO out 6 slots = C(6,3) x 3! = 120
15. How many permutations are there of the letters in the word TOTTERER if: 8 letters, 3T, 2E, 2R
a. The E’s are adjacent? Count 2E as one = 7! ÷ (3! x 2!) = 420
b. The E’s are not adjacent? 8! ÷ (3! x 2! x 2!) – 7! ÷ (3! x 2!) = 1260
c. The E’s are adjacent and the R’s are adjacent? 6! ÷ 3!
d. The T’s occur in the 2nd, 4th, and 5th position? 5! ÷ 2! ÷ 2!
16. How many arrangements can be formed using the letters in the word SHRINKS are there if the S
must be directly in front of the H? treat SH as one 6! ÷ 2
17. In how many ways can eight test questions be arranged so that the easiest and the most difficult
questions do not immediately follow one another? 8! – 2x7!
18. How many 4‐digit numbers are there with the following restrictions:
a. Using the digits 1 to 8.
P(8,4) if no repletion; 8x8x8x8 if repetition
b. Using the digits 0 to 7.
0 can’t be the first 7 x P(7,3) or 7x8x8x8
c. Using only odd digits.
P(5,4) or 5x5x5x5
d. The number is odd.
9x10x10x5
e. The number is even.
9x10x10x5
19. How many 3‐digit numbers can be formed using only the digits 1 to 7 if the number 2 must be
included?
3 locations for number 2 & 2 from the 6 left = 3 x P(6,2) = 90
20. The Greek alphabet has 24 characters. How many different Greek‐letter fraternity names can be
formed using two or three letters? (repetition is permitted)
24x24 + 24x24x24 = 14400
28. Evaluate each of the following.
(a) C(8, 3) 56
(b) 2C1
(d) C(21, 13) 203490
(e)
29. Solve for n.
(a) C(n, 2) = 66
12
2
5C2 10
(b) 6P(n + 1, 3) = 48C(n, 4)
(c)
10C4 210
no sol
30. In how many ways can a child select three different ice cream flavours from nine different ice
cream flavours? 9C3 = 84
31. How many hands of six cards can be selected from a standard deck of 52 cards?
52C6 = 20358520
32. In how many ways can a group of five people be chosen from seven couples (each of which has one
male and one female) to form a club, given each of the following conditions?
(a) All are equally eligible for the club. 14C5
(b) The club must include two females and three males. (7C2)(7C3)
33.If a closed polygon has 20 diagonals, how many sides does it have? 8
34. ProvethatC(n+1,r)=C(n,r)+C(n,r–1).
35 A gardener wants to plant a line of six trees. He has available 10 cedars, 7 oaks, 8 maples, 9 willows
and 12 pines. In how many ways he can arrange his line of trees? 56
Ex. 3: How many 4 letter-words can be created out of the word HELLO.
All different 4! = 24
With 2 L: (slots for 2 L)(choose 2 from 3 others) = C(4,2)P(3,2)= 36
Total = 24 + 36 = 50
Ex 5: How many 4 letter-words can be created out of MISSISSIPPI.
4S 4I 2P 1M
Direct Reasoning:
all the same only S or I = 2
3 the same and 1 = (2 for 3 the same)(3 left)(4 slots for single) = 24
2 the same and 2 different = (3 for 2 the same)xP(3,2)xC(4,2) = 108
2 sets the same = 3 choices for the first, 2 choices for the second P(3,2)xC(4,2)/2= 18
(because they are symmetric here we have to divide by 2)
all different = 4! = 24
total = 2 + 24 + 108 + 18 + 24 = 176
Indirect Reasoning:
If we have 4 for each total 4x4x4x4 = 256
Impossible cases:
All P or all M = 2 cases
3 M and one other 3x4 = 12 cases
3 P and one other = 12 cases
2 M and 2 the same = 3xC(4,2) = 18
2 M and 2 different = P(3,2)C(4,2) = 36
256 – (2+ 12 + 12 + 18 +36) = 176
Poker game hands
Hand
Royal flush
Description
Number of combinations
A, K, Q, J, 10 of a suit
4
Straight flush
Five consecutive cards of one suit, without
any ace
9 x 4 = 36
Four of a kind
4 of the same kind and one is optional
13 x C(4,4) x C(48,1)
= 624
Three of a kind and a pair
13 x C(4,3) x 12 x C(4,2)
= 3744
5 cards of the same suit (not consecutive)
4 x C(13,5) – 4 – 36
= 5108
Straight
Cards of the consecutive value (not the
same suit)
40x4x4x4x4 – (4 + 36)
= 10200
Three of a kind
Three cards of the same face value; the
other two are different value
13 x C(4,3) x48 x44 /2
= 54912
Two set of the same face value
13xC(4,2) x12C(4,2) x 44/2
= 123552
Only two card with the same face value
(13 x C(4,2) x 48 x 44 x 40) /3!
= 1098240
All single face value
52x48x44x40x36/5! – (Flush +
3 kind + straight)
= 1302540
C(52,5)
2598960
Full house
Flush
Two pairs
One pair
No pair
Total