QS 026 CHAPTER 9 PERMUTATIONS and COMBINATIONS PERLIS

QS 026 CHAPTER 9 PERMUTATIONS and COMBINATIONS PERLIS MATRICULATION COLLEGE

Permutations of a set of objects A permutation of a set of objects is any arrangement of those objects in a definite order. (order is important). For example, if A={a, b, c, d} , then ab two-element permutation of A, acd three-element permutation of A, adcb four-element permutation of A.

The order in which objects are arranged is important. For example, n ab and ba are considered different twoelement permutations n n abc and cba are distinct three-element permutations, and abcd and cbad are different four-element permutations.

For another example the six permutations of ABC are the six different arrangements of ABC. These are ABC ACB BAC BCA CAB CBA

If there are 4 ways from Johor to Penang and 2 ways from Penang to Pulau Langkawi, how many ways can we go for a journey from Johor to Pulau Langkawi through Penang.

Bus Taxi Johor Train Van Flight Penang Ferry P. Langkawi

The number of permutations can be calculated using the multiplication principle. Multiplication Principle If there are m ways for an event to occur and n ways for another event to occur, then there are m x n ways for the two events to occur.

Example A coin and a dice are tossed together. How many different outcomes are possible ? Solution The coin has two possible outcomes (head, H and Tail, T) and the dice has 6 possible outcomes. The number of different possible outcomes is 12 2 x ___ 6 ___ = ____ The possible outcomes are (H, 1) (H, 2) (H, 3) (H, 4) (H, 5) (H, 6) (T, 1) (T, 2) (T, 3) (T, 4) (T, 5) (T, 6)

Die 1 Coin 2 3 4 5 6 Head (H) (H, 1) (H , 2) (H, 3) (H, 4) (H, 5) (H, 6) Tail (T) (T, 1) (T, 2) (T, 3) (T, 4) (T, 5) (T, 6)

Example A shop stocks T-shirts in four sizes : small, medium and large. They are available in four colours; black , red , yellow and green. If the sizes are denoted by S, M and L and the colours are denoted by B, R, Y and G make a list of all the different labels needed to distinguish the T-shirts and find the number of different labels. Solution SB MY SR MG SY LB SG LR MB LY MR LG The number of different labels is 3 x 4 = 12

Permutations of n objects We will now consider the method for finding a number of permutations on the letters A, B and C using the multiplication principle. How many arrangements of the letters A, B and C are there? Let us consider the number of ways of arranging n letter. If we have 1 letter, there is just one arrangement. Example : A

If we have 2 letters, there are two different arrangements. Example : AB and BA If we have 3 letters, the different arrangements are : there are three ways of choosing the first letter.

When the first letter has been chosen, there are two letters from which to choose the second; and the possible ways of choosing the first two letters are: there are two ways of choosing the second letter

i. e. for each of the three ways of choosing the first letter, there are two ways of choosing the second letter. Hence there are 3 x 2 ways of choosing the first two letters. Having chosen the first two letters, there is only one choice for the third letter, i. e. for each of the 3 x 2 ways of choosing the first two letters, there is only one possibility for the third letter. Hence there are 3 x 2 x 1 ways of arranging the three letters A, B and C. Note : if repetition are allowed, we can choose from all 3 digits for each digit of the number. A digit can be used more than once.

How many different ways of arranging 3 digit numbers from digits 5 and 6 ? 5 5 5 6 6 5 6 5 6 there will be 8 different ways, which is found from 2 2 2

n How many different ways do you think there are of arranging 4 letters? You should able to see, there will be 24 different ways, which is found from 4 x 3 x 2 x 1. If there are 500 different objects, the number of ways would be 500 x 499 x 498 x … x 3 x 2 x 1. This is tedious to write, so we use the notation 500! ( factorial 500 ).

Example List the set of all permutations of the symbols P, Q and R when they are taken 3 at a time Solution PQR, PRQ, QPR, QRP, RPQ, RQP

In general, Number of permutations of n different objects taken all at a time without repetition n P n = n x (n – 1) x(n – 2) x … x 2 x 1 = n! n Notes : P n = n! n! means the products of all the integers from 1 to n inclusive and is called ‘n factorial’.

Example How many three-digit numbers can be made from the integers 2, 3, 4 ? Solution n=3 n n = 3!3 x 2 x 1=6 The number of arrangements is 6.

Example In how many ways can ten instructors be assigned to ten sections of a course in mathematics? Solution Substituting n = 10 we get = = 10 ! = 3, 628, 800 ways

Example Three people, Aishah, Badrul and Daniel must be scheduled for job interviews. In how many different orders can this be done? Solution n=3 So there are 3! = 6 possible orders for the interviews.

Example How many different numbers can be formed from the digits 5, 6, 7 and 8 i) if no repetitions are allowed n=4 np = 4! = 24 n 4 ii) if the first digit must be 7. 1 p X 3 p = 1! X 3! = 6 1 3