1
PERMUTATIONS and COMBINATIONS
Combinations – order doesn’t count
1. A Champions League group consists of four teams, Ajax, Barcelona, Celtic, and Dortmund.
Two of these teams qualify from the group.
Write down the different combinations of teams that can qualify.
4
AB, AC, AD,
BC, BD, CD
4∙3
4
=6
𝐶𝐶2 = � � =
2
2
1a. A certain tournament does not require the coach to state the ranking of the singles players.
Instead, the coach only needs to submit 3 names for the single players. In how many ways this
can be done out of 8 players?
8
8∙7∙6
8
𝐶𝐶3 = � � =
= 56
3
3!
1b. A sports committee at the local hospital consists of 5 members. A new committee is to be
elected, of which 3 members must be women and two members must be men. How many
different committees can be formed if there were originally 5 women and 4 men to select from?
Permutations - order counts
4 5
� � � � = 60
2 3
2. A Champions League group consists of four teams, Ajax, Barcelona, Celtic, and Dortmund.
Write down the different ways the group can finish when they have played all their matches.
ABCD,
BACD,
CABD,
DABC,
ABDC,
BADC,
CADB,
DACB,
ACBD,
BCAD,
CBAD,
DBAC,
ACDB,
BCDA,
CBDA,
DBCA,
ADBC,ADCB
BDAC, BDCA
CDAB, CDBA
DCAB, DCBA
4! = 24
Permutations - selections
3. A Champions League group consists of four teams, Ajax, Barcelona, Celtic, and Dortmund.
Two of these teams qualify from the group, one as winners and the others as runners-up. Write
down the possible ways in which teams can qualify as winners and runners-up.
AB, BA, AC, CA, AD, DA, BC, CB, BD, DB, CD, DC
4
𝑃𝑃2 =
4!
2!
= 12
3a. The IASAS tennis team consists of 8 members. In how many ways can the coach select the
1st, 2nd and 3rd singles players?
8
𝑃𝑃3 =
8!
= 336 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤
5!
2
Restrictions 1
4. 6 people (A,B,C,D,E,F) are to line up. Two of the people (A,B) must be next to each other.
How many arrangements can be made?
Think of the two people as one letter, so arrangements can be:
AB,C,D,E,F C,AB,D,E,F etc.
but for every arrangement you can reverse A and B,
BA,C,D,E,F C,BA,D,E,F etc.
5! = 120
5!x2! = 240
5. 6 people (A,B,C,D,E,F) are to line up. Three of the people (A,B,C) must be next to each other.
How many arrangements can be made?
4!x3!=144
Restrictions 2
6. 8 athletes are to be lined up for a race. 2 of athletes are from Zambia, and one each from
Angola, Botswana, Cameroon, DR Congo, Egypt and Ghana. The two Zambian athletes are not
allowed to be next to each other.
How many ways can the athletes line up.
Without a restriction: 8!=40320
Take away the restrictions: 7!x2 =10080
40320-1008=30240
7. 8 athletes are to be lined up for a race. 2 of athletes are from Zambia, and one each from
Angola, Botswana, Cameroon, DR Congo, Egypt and Ghana. The two Zambian athletes are not
allowed to be next to each other, neither are the Ghanian or Congolese athletes allowed to be
next to each other.
How many ways can the athletes line up.
8! – (7!x2) – (7!x2) +(6!x2x2) = 20160+(6!x2x2) = 23040
Repetitions
8. 10 books are to be lined up on the shelf
a. How many ways can the books be lined up.
b. If 5 of the books are identical math books and 2 are identical Science books and 3 are
identical English books, how many ways can they be lined up?
a) Without a repetition: 10!=36288000
b) With repetition: permutation of identical books is nothing new
9. How unique 4 letter “words” can be made from: COCA-COLA
8!
P4
= 4! = 70
3! 2! 2! 3! 2! 2!
8
10!
5!2!3!
=2520
3
Choosing people and restrictions
10. 6 people are to be chosen for a new committee from 8 males and 8 females.
How many different ways can the committee be chosen if,
a) there are no restrictions on who is chosen,
b) there must be equal males and females on the committee
c) the current chairperson must be re-elected to the committee?
d) there must be at least 4 females on the committee,
𝑎𝑎)
16
𝑐𝑐)
�
𝐶𝐶6 = 8008
8
𝑏𝑏)
𝐶𝐶3 × 8𝐶𝐶3 = 3136
15
� = 3003
5
8
𝑑𝑑)
𝐶𝐶4 × 8𝐶𝐶2 + 8𝐶𝐶5 × 8𝐶𝐶1 + 8𝐶𝐶6 = 2436
IB PROBLEMS
11.
There are 10 seats in a row in a waiting room. There are six people in the room.
[6]
(a) In how many different ways can they be seated?
(b) In the group of six people, there are three sisters who must sit next to each other.
In how many different ways can the group be seated?
(a) A recognition of a permutation of six from ten in words or symbols
Total number of ways =151200
(b) Total number of ways = 8 x 3! x 7 x 6 x 5 = 10 080
12. Twelve people travel in three cars, with four people in each car. Each car is driven by its
owner. Find the number of ways in which the remaining nine people may be allocated to the cars.
(The arrangement of people within a particular car is not relevant).
[6]
9
The first car can be filled in
The second car can be filled in
The third car can be filled in
Number of combinations =
9
𝐶𝐶3
3
ways.
6
𝐶𝐶3
𝐶𝐶3
6
ways.
ways.
𝐶𝐶3 × 𝐶𝐶3 × 3𝐶𝐶3 = 84 × 20 × 1 = 1680
4
13. In how many ways can six different coins be divided between two students so that each
student receives at least one coin?
[3]
The first student can receive x coins in
�𝑥𝑥6�
[The second student then receives the rest.]
ways, 1 ≤ 𝑥𝑥 ≤ 5
Therefore, the number of ways = �61� + �62� + �63� + �64� + �65� = 26 − 2 = 62
𝑛𝑛
𝑛𝑛
� � � = 2𝑛𝑛
𝑖𝑖
𝑖𝑖=0
14.
𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 𝑏𝑏𝑏𝑏 𝑚𝑚𝑚𝑚𝑚𝑚ℎ 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖:
→
5
6
6
6
� � � = 26 − � � − � � = 26 − 2
𝑖𝑖
0
6
𝑖𝑖=1
How many four-digit numbers are there which contain at least one digit 3?
[3]
The total number of four-digit numbers = 9 x 10 x 10 x 10 = 9000.
The number of four-digit numbers which do not contain a digit 3 = 8 x 9 x 9 x 9 = 5832.
Thus, the number of four-digit numbers which contain at least one digit 3 is 9000 – 5832 = 3168.
15. A committee of four children is chosen from eight children. The two oldest children cannot
both be chosen. Find the number of ways the committee may be chosen.
[6]
METHOD 1
Consider the group as two groups – one group of the two oldest and one group of the rest
Either one of the two oldest is chosen or neither is chosen
Then the number of ways to choose the committee is
METHOD 2
The number of ways to choose a committee of 4 minus the number of ways to have both the
oldest
16. From a group of 12 people, 8 are chosen to serve on a committee.
(a) In how many different ways can the committee be chosen?
(b) One of the 12 people is called Sameer. What is the probability that he will be on the
committee?
(c) Among the 12 people there is one married couple. Find the probability that both partners
will be chosen.
(d) Find the probability that the three oldest people will be chosen.
5
(a) This is the combination
�12
�
8
= 495, which acts as the total possibility space.
(b) Since Sameer must be on the committee, we need to choose 7 people from 11. This can be
done in
11
� � = 330 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤.
7
2
330
Hence the probability that Sameer will be on the committee is
=
3
495
(c) As in part (b) we know that the married couple will be on the committee and need to choose
six people from ten. This can be done in
10
� � = 210 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤.
6
14
210
Hence the probability that the married couple will be on the committee is
=
33
495
(d) Since the three oldest people have been chosen, we need to choose 5 people from 9
remaining.
9
� � = 126 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤.
5
14
126
Hence the probability that three oldest people will be on the committee is
=
55
495
17. A committee of 3 men and 2 women is to be chosen from 7 men and 5 women. Within 12
people there is a husband and wife. In how many cases can the committee be chosen if it must
contain either the wife or the husband but not both?
6 4
6 4
� �� � + � �� �
2 2
3 1
18. Ten students in a class are divided into two groups of five to play in a five-a-side soccer
tournament. In how many ways can the two teams of five be selected?
This arrears to be very similar to the example before, but there is a subtle difference. The number of
10
ways of selecting a team of five is � � = 252. Let us imagine that the chosen team is ABCDE.
5
Hence the other team would automatically be FGHIJ. However another possible combination of a
team of five would be FGHIJ and this would then automatically select the other team to be
ABCDE. In other words the calculation picks each pair of teams twice. Hence the actual number of
teams is
10
𝐶𝐶5
= 126
2
6
18. Four letters are picked from the word EXAMPLES.
(a) How many different arrangements are there of the four letters.
(b) What is the probability that the arrangement of four letters will not contain a letter E?
(c) What is the probability that the arrangements will contain both of the letter E’s?
(d) Given that the arrangements contains both of the letter E’s, what is the probability that the two letter E’s will
be separated?
𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐:
Number of permutations that do not contain E is: 6𝑃𝑃4 = 360
Number of permutations that contain one E is: 4 x 6𝑃𝑃3 = 480
E first and 6𝑃𝑃3 = 120 permutations, E second and 6𝑃𝑃3 = 120 permutations,
E third and 6𝑃𝑃3 = 120 permutations, E fourth and 6𝑃𝑃3 = 120 permutations
Number of permutations that contain two E’s is: 5 × 6𝑃𝑃2 = 5 × 30 = 150
EE can be first and second, first and third, first and fourth, second and third and third and fourth
(a) total possible number of arrangements is 360 + 480 + 150 = 870
(b) P(E’) = 360/870 = 12/29
(c) P(2E) = 150/870 = 5/29
(d) Number of arrangements with 2E = 150,
Number of permutations that contain two separated E’s is: 2 × 6𝑃𝑃2 = 2 × 30 = 60
P(two separated E)= 60/150 = 2/5
19. Solve
a.
n:
2n
P3 = 2 ( n P4 )
a.
n≥ 4
because of
 n + 1  n   2 n 

 −  =
 
 3   2  2 
3
b.
nP4
2n(2n-1)(2n-2) – 2(n)(n-1)(n-2)(n-3) = 0
(2n-1)(2)(n-1) – (n-1)(n-2)(n-3) = 0
(n-1)[2(2n-1) – (n-2)(n-3)] = 0
n≥ 4
2(2n-1) – (n-2)(n-3) = 0
(n-1)(n-8) = 0
b.
3
2
so n– 1≠ 0
4n – 2 – n2 + 5 n – 6 = 0
n=8
4
20. Three identical door prizes are to be given to three lucky people in a crowd of 100. In how
many ways can this be done?
100
𝐶𝐶3 = 161700
7
21. Baskin Robbins has 20 flavours of ice cream and 11 flavours of sherbet. In how many ways
could you select
b. A scoop of ice cream and then a scoop of sherbet?
c. A scoop of ice cream or a scoop of sherbet?
a.
b.
20 x 11 = 220
20 + 11 = 31
22. There are 3 roads from town A to town B, 5 roads from town B to town C and 4 roads from
town C to town D. How many different ways are there to go from A to D via B and C? How many
different round trips are possible?
a. 3 x 4 x 5 = 60
b. 602 = 3600
23. A teacher must pick 3 high school students from a class of 30 to prepare and serve food at
the junior high school picnic. How many choices are possible?
30
𝐶𝐶3 = 4060
24. How many ways can 8 jackets of different styles be hung, on a straight bar? On a circular
track?
a. 8! = 40320
b. 7! = 5040
25. A basketball team has 4 guards, 5 forwards and 3 centers on it’s roster. In how many ways
can a team be created with 1 right guard, 1 left guard, 1 right forward, 1 left forward and 1
center?
4 x 3 x 5 x 4 x 3 = 720
26. A bookshelf has space for 5 books. If 7 books are available, how many different
arrangements can be made on the bookshelf?
7
𝑃𝑃5 = 2520
27. A hockey team consists of 10 Canadians, 2 Americans and 6 Europeans they can each play
all positions, the coach selects 6 players for the starting line up. How many ways can the coach
select 4 Canadians, 1 American and 1 European?
10
𝐶𝐶4 2𝐶𝐶1 6𝐶𝐶1 = 2520
28. A town council consists of 8 members including the mayor
d. How many different committees of 4 can be chosen from this council?
e. How many of these committees include the mayor?
f. How many of these committees do not include the mayor?
a. 8𝐶𝐶4 = 70
b. 7𝐶𝐶3 = 35
c. 70 – 35 = 35
8
29.
A biased die with four faces is used in a game. A player pays 10 counters to roll the
die. The table below shows the possible scores on the die, the probability of each score
and the number of counters the player receives in return for each score.
Score
1
2
3
4
Probability
1
2
1
5
1
5
1
10
Number of counters player
receives
4
5
15
n
Find the value of n in order for the player to get an expected return of 9 counters per
roll. [4]
Let X be the number of counters the player receives in return.
E(X) = ∑p(x) × x = 9
⇔
1
n=3
10
⇔
⇔

 1
 1  1
1
 × 4  +  × 5  +  × 15  +  × n  = 9

  10
 5  5
2
n = 30