Page 9 - Pra U STPM 2022 Penggal 3 - Maths (T)

Mathematics Semester 3 STPM Chapter 2 Probability
Combinations

In dealing with permutations the order of the elements does matter. For instance, in Example 6 the coach’s
selection of the first and second singles from the 5 players, the arrangement would be different if we exchanged
the positions of the 2 selected players.

Now, let us consider different type of selections. If the coach wished to pick a doubles from the 5 players,
the order of choosing the first and the second player does not really matter. We begin with how many ways
we can choose 2 players from a group of 5 players and arrange them in order. It was found to be P = 5!
5
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2 3!
= 20 ways. Assume the players are designated by A, B, C, D or E. For a group of 2 players A and B, we
have two arrangements, AB and BA. Note that the doubles consisting of AB is the same doubles formed by 2
BA. This implies that within the 20 ordered arrangements, the order of the groups of 2 players should be
disregarded. Consequently, we divide the ordered arrangements of 5 players taken 2 at a time, P , by the
5

2
number of ways of 2 players arranged between them, i.e. 2!.
5 P
This gives 2 = 5!
2! 3! 2!
5 × 4
=
2
= 10 doubles

Generally, we could extend the above analysis by choosing r elements from a set of n elements without regard
r 1 2
to order. This type of selections is classified as combination and is denoted by C or n . The number of
n
r
possible combinations, C , is calculated by the second permutations formula divided by r!. Thus
n

r
n P
C = =
n r n!
r r! (n – r)! r! Choosing without
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Example 9

How many different ways can you choose 4 kings from a standard deck of 52 playing cards?

Solution: This problem involves combinations, because the order of the cards does not
matter. We select 4 kings from 4 available kings. So,
4!
4 C =
4 (4 – 4)! 4!
= 1

There is only 1 way you could select 4 kings.

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02 STPM Math(T) T3.indd 77 28/10/2021 10:21 AM