Page 8 - Pra U STPM 2022 Penggal 3 - Maths (T)

Mathematics Semester 3 STPM Chapter 2 Probability

Example 6

A badminton team has 5 players. How many ways can a coach select the first and second singles?
Solution: Here we are choosing 2 from 5 and arranging them in order. So, the number of
ways are
5 P = 5 × 4 = 20
2
Thus, there are 20 ways for the coach to select the first and second singles.
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So far we have considered permutations of distinct elements. If the letters B and C are both equal to X, then
2 the 6 permutations of the letters A, B and C becomes AXX, AXX, XXA, XAX, XAX, XXA, of which only 3

are different. Thus, with 3 letters, 2 being the same, there are 3! = 3 distinct permutations.
2!

In general, the number of distinct permutations of n elements of which n are of one kind, n of a second
1
2
kind, …, n of a kth kind is given by the formula
k
n!
n ! n ! … n !
1 2 k
Example 7
Find the number of different ways to arrange 2 yellow, 3 red and 4 green bulbs in a string of Christmas
tree lights with 9 sockets.

Solution: The total number of different arrangements is
9!
= 1260
2! 3! 4!

Example 8

Find the number of arrangements that can be formed from all the letters of the word MATHEMATICS, if
(a) there is no restrictions,
(b) the two M’s are separated.

Solution: For the word MATHEMATICS, there are 11 letters with 2 A’s, 2 M’s, 2 T’s, 1 H,
1 E, 1 I, 1 C and 1 S.
(a) If there is no restrictions, the total number of different arrangements
= 11! = 4 989 600
2! 2! 2! 1! 1! 1! 1! 1!

(b) If the two M’s are together, they can be considered as a single letter and
the total number of different arrangements
= 10! = 907 200
2! 2! 1! 1! 1! 1! 1! 1!
Thus, if the two M’s are separated, the total number of different arrangements
= 4 989 600 – 907 200
= 4 082 400
Note: This is an example of permutation with restriction.

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02 STPM Math(T) T3.indd 76 28/10/2021 10:21 AM