1 PERMUTATIONS and COMBINATIONS - Uplift Education

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.

AB, AC, AD, BC, BD, CD

42

=

42

=

43 2

=

6

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?

83

=

83

=

8

7 3!

6

=

56

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?

45 2 3 = 60

Permutations - order counts 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, ABDC, ACBD, ACDB, ADBC,ADCB BACD, BADC, BCAD, BCDA, BDAC, BDCA CABD, CADB, CBAD, CBDA, CDAB, CDBA DABC, DACB, DBAC, DBCA, 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 2

=

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 3

=

8! 5!

=

336

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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

10!

b) With repetition: permutation of identical books is nothing new

=2520

5!2!3!

9. How unique 4 letter "words" can be made from: COCA-COLA

8P4 =

8! 4!

= 70

3! 2! 2! 3! 2! 2!

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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,

) 166 = 8008 ) 83 ? 83 = 3136 ) 155 = 3003 ) 84 ? 82 + 85 ? 81 + 86 = 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]

The first car can be filled in 93 ways. The second car can be filled in 63 ways. The third car can be filled in 33 ways.

Number of combinations = 93 ? 63 ? 33 = 84 ? 20 ? 1 = 1680

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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 6 ways, 1 5

[The second student then receives the rest.]

Therefore, the number of ways = 61 + 62 + 63 + 64 + 65 = 26 - 2 = 62

:

=

2

=0

5

6

=

26

-

6 0

-

6 6

=

26 - 2

=1

14. 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.

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(a) This is the combination 182 = 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 7 = 330 .

Hence the probability that Sameer will be on the committee is 330 = 2

495 3 (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 6 = 210 .

Hence the probability that the married couple will be on the committee is 210 = 14

495 33

(d) Since the three oldest people have been chosen, we need to choose 5 people from 9 remaining.

9 5 = 126 .

Hence the probability that three oldest people will be on the committee is 126 = 14

495 55

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?

64 64 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

ways of selecting a team of five is 150 = 252. Let us imagine that the chosen team is ABCDE.

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

105 = 126 2

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