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PERMUTATIONS AND COMBINATIONS
Multiplication Rule If one event can occur in m ways, a second event in n ways and a third event in r , then the three events can occur in m x n x r ways. Example: In how many ways can Artemis select one top, one skirt and one cap if she has 6 tops, 5 skirts and 4 caps from which to choose an outfit.? Solution: Ways= 6 x 5 x 4 = 120 ways Repetition of an Event If one event with n outcomes occurs r times with repetition allowed, then the number of ordered arrangements is nr. Example 1: What is the number of arrangements if a die is rolled (a) 2 times? 6 x 6 = 36 (b) 3 times? 6 x 6 x 6 = 216 (c) r times? 6 x 6 x 6 x … = 6r Example 2: (a) How many different cars number plates are possible with 3 letters followed by 3 digits? Solution: 26 x 26 x 26 x 10 x 10 x 10 = 26^3 x 10^3 = 17,576, (b) How many of these number plates begin with ABC? Solution: 1 x 1 x 1 x 10 x 10 x 10 = 10^3 = 1000 (c) If a plate is chosen at random, what is the probability that it begins with ABC?
Solution:
3 26 3 × 10
3 =^
3 =^
Factorial Representation n! = n(n-1)(n-2)…3 x 2 x 1 For example, 5! = 5 x 4 x 3 x 2 x 1 Note: 0! = 1 Example: (a) In how many ways can 6 people be arranged in a row? Solution: 6! = 6 x 5 x 4 x 3 x 2 x 1 = 720 ways (b) How many arrangements are possible if only 3 of them are chosen? Solution: 6 x 5 x 4 = 120 Arrangements or Permutations Distinctly ordered sets are called arrangements or permutations. The number of permutations of n objects taken r at a time is given by: nPr =^ n! ( n − r )! where n = number of objects r = number of positions Example 1: A math’s debating team consist of 4 speakers. (a) In how many ways can all 4 speakers be arranged in a row for a photo? Solution: (^) 4 P 4 =
4 × 3 × 2 × 1
(b) How many ways can the captain and vice-captain be chosen?
Solution= 2 x 8 P 8 = 2 x 8! = 80640 Arrangements with Repetitions If we have n elements of which x are alike of one kind, y are alike of another kind, z are alike of another kind, then the number of ordered selections or permutations is given by: P = n! x! y! z! Example 1: How many different arrangements of the word PARRAMATTA are possible? Solution: 10 letters but note repetition (4 A’s, 2 R’s, 2 T’s) P No. of arrangements =
A A A A No. of arrangements = 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 4 × 3 × 2 × 1 × 2 × 1 × 2 × 1
10 × 9 × 8 × 7 × 6 × 5
R R
M
T T
Arrangements with Restrictions Example 1: How many different arrangements of the word REMAND are possible if: (a) there are no restrictions? Solution: (^) 6 P 6 = 6! = 720
(b) they begin with RE? Solution: R E _ _ _ _ = 4 P 4 = 4! = 24 (c) they do not begin with RE? Solution: Total – (b) = 6! – 4! = 720 – 24 = 696 (d) they have RE together in order? Solution: (RE) _ _ _ _ = 5 P 5 = 5! = 120 (e) they have REM together in any order? Solution: (REM) _ _ _ = 3 P 3 x 4 P 4 = 3!4! = 144 (f) R, E and M are not to be together? Solution: Total – (e) = 720 – 144 = 576 Example 2: There are 6 boys who enter a boat with 8 seats, 4 on each side. In how many ways can (a) they sit anywhere? Solution: (^) 8 P 6 =
8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
2 × 1
(b) two boys A and B sit on the port side and another boy W sit on the starboard side? Solution: A & B = 4 P 2 W = 4 P 1 Others = 5 P 3 Total = 4 P 2 x 4 P 1 x 5 P 3 =
24 × 24 × 120
2 × 6 × 2
Example 3: From the digits 2, 3, 4, 5, 6 (a) how many numbers greater than 4000 can be formed? Solution: 5 digits (any) = 5 P 5 4 digits (must start with digit ≥^ 4 = 3 P 1 x 4 P 3
(b) men and women alternate Solution: (6-1)! x 6! = 5!6! = 120 x 720 = 86400 (c) Ted and Carol must sit together Solution: (TC) and other 10 = 2! x 10! = 7,257, (d) Bob, Ted and Carol must sit together Solution: (BTC) and other 9 = 3! x 9! = 2,177, (e) neither Bob nor Carol can sit next to Ted Solution: Seat 2 of the other 9 people next to Ted in (9 x 8) ways or 9 P 2 Solution: Then sit the remaining 9 people (including Bob and Carol) in 9! Ways Ways = (9 x 8) x 9! or 9 P 2 x 9! = 72 x 362880 = 26,127, Example 2: In how many ways can 8 differently colored beads be threaded on a string? Solution: As necklace can be turned over, clockwise and anti-clockwise arrangements are the same = (8-1)! ÷ 2 = 7! ÷ 2 = 2520 Unordered Selections The number of different combinations (i.e. unordered sets) or r objects from n distinct objects is represented by: No. of combinations = number of permutations arrangements of r objects and is denoted by nCr =^ n P r r!
n! r! ( n − r )!
Combinations Example 1: How many ways can a basketball team of 5 players be chosen from 8 players? Solution: (^) 8 C 5 =
8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
5 × 4 × 3 × 2 × 1 × 3 × 2 × 1
Example 2: A committee of 5 people is to be chosen from a group of 6 men and 4 women. How many committees are possible if (a) there are no restrictions? Solution: (^) 10 C 5 =
10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
5 × 4 × 3 × 2 × 1 × 5 × 4 × 3 × 2 × 1
(b) one particular person must be chosen on the committee? Solution: 1 x 9 C 4 =^1 ×^
9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
4 × 3 × 2 × 1 × 5 × 4 × 3 × 2 × 1
(c) one particular woman must be excluded from the committee? Solution: (^) 9 C 5 =
9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
5 × 4 × 3 × 2 × 1 × 4 × 3 × 2 × 1
(d) there are to be 3 men and 2 women? Solution: Men and women = (^) 6 C 3 x (^) 4 C 2 = 6_!_ 4_!_ 3_!_ ( 6 − 3 )! 2_!_ ( 4 − 2 )!
6 × 5 × 4 × 3 × 2 × 1 × 4 × 3 × 2 × 1
3 × 2 × 1 × 3 × 2 × 1 × 2 × 1 × 2 × 1
(e) there are to be men only? Solution: (^) 6 C 5 =
6 × 5 × 4 × 3 × 2 × 1
5 × 4 × 3 × 2 × 1
(f) there is to be a majority of women? Solution: 3 women and 2 men or 4 women and 1 man = 4 C 3 x 6 C 2 + 4 C 4 x 6 C 1
Solution: (^) 13 C 5 =
13 × 12 × 11 × 10 × 9 × 8 × …× 1
5 × 4 × 3 × 2 × 1 × 8 × … × 1
d) all the same color? Solution: red or black 26 C 5 + 26 C 5 = 2 x 26 C 5 26 C 5 =^
26 × 25 × 24 × 23 × 22 × 21 × … × 1
5 × 4 × 3 × 2 × 1 × 21 × … × 1
26 C 5 +^26 C 5 = 2 x^26 C 5 = 2 x 65780 = 131560 e) four of the same kind? Solution: (^) 4 C 4 x 48 C 1 x 13 4 C 4 =^
48 C 1 =^
48 × 47 ×… × 1
1 × 47 × … × 1
4 C 4 x^48 C 1 x 13 = 1 x 48 x 13 = 624 f) 3 Aces and 2 Kings? Solution: (^) 4 C 3 x 4 C 2 4 C 3 =^
4 × 3 × 2 × 1
3 × 2 × 1 × 1
4 C 2 =^
4 × 3 × 2 × 1
2 × 1 × 2 × 1
4 C 3 x^4 C 2 = 4 x 6 = 24 Further Permutations and Combinations Example 1: If 4 Math books are selected from 6 different Math books and 3 English books are chosen from 5 different English books, how many ways can the seven books be arranged on a shelf
(a) if there are no restrictions? Solution: (^) 6 C 4 x 5 C 3 x 7! 6 C 4 =^
6 × 5 × 4 × 3 × 2 × 1
4 × 3 × 2 × 1 × 2 × 1
5 C 3 =^
5 × 4 × 3 × 2 × 1
3 × 2 × 1 × 2 × 1
6 C 4 x^5 C 3 x 7! = 15 x 10 x 5040 = 756000 (b) If the 4 Math books remain together? Solution: (MMMM) _ _ _ 6 P 4 x 5 C 3 x 4! 6 P 4 =^
6 × 5 × 4 × 3 × 2 × 1
2 × 1
5 C 3 =^
5 × 4 × 3 × 2 × 1
3 × 2 × 1 × 2 × 1
6 P 4 x^5 C 3 x 4! = 360 x 10 x 24 = 86400 or ( 6 C 4 x 4!) x 5 C 3 x 4! 6 C 4 =^
6 × 5 × 4 × 3 × 2 × 1
4 × 3 × 2 × 1 × 2 × 1
5 C 3 =^
5 × 4 × 3 × 2 × 1
3 × 2 × 1 × 2 × 1
( 6 C 4 x 4!) x 5 C 3 x 4! = (15 x 24) x 10 x 24 = 86400 (c) a Math book is at the beginning of the shelf Solution: M _ _ _ _ _ _ 6 x 5 C 3 x 5 C 3 x 6!
SYLLABUS = 10080 permutations (ii) If a word is chosen at random, find the probability that the word: a) contains the two S’s together Solution: (SS) _ _ _ _ _ _ Words =
= (^2520) Probability =
b) begins and ends with L Words =
= (^360) Probability =
