Page 10 - Pra U STPM 2022 Penggal 3 - Maths (T)

Mathematics Semester 3 STPM Chapter 2 Probability

Example 10

From 5 men and 3 women, find the number of committees of 5 that can be formed with 3 men and 2
women.
Solution: The number of ways of selecting 3 men from 5 is

5 C = 5!
3 (5 – 3)! 3!
= 10
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The number of ways of selecting 2 women from 3 is
2 3 C = 3!
2 (3 – 2)! 2!
= 3

Thus, the required number of committees that can be formed = 10 × 3
= 30
Note: We apply the multiplication principle in this example.

Note: A commonly raised question is, “When do we use permutation and when do we use combination?”
To answer this we must first realise that objects are drawn from the lot without replacement. Thus,
the total number of ways is P if the order in which the objects are drawn is important, and C if
n
n
r
r
it is not important.
Example 11

Consider the collection of objects consisting of the six letters a, b, c, d, e, f. Write the answers in factorial
notations.
(a) Find the number of three-letter permutations.
(b) Find the number of three-letter combinations.
(c) Which is greater, the number of permutations or the number of combinations?
(d) Find an equation relating parts (a) and (b).
6!
Solution: (a) The number of three-letter permutations is, P =
6
3 (6 – 3)!
= 6!
3! 6!
(b) The number of three-letter combinations is, C =
6
3 (6 – 3)! 3!
6!
=
3! 3!
(c) From parts (a) and (b), it is obvious that the number of permutations is
greater than the number of combinations
6! 6 P
6
(d) The number of three-letter combinations is, C = = 3
3 3! 3! 3!
6 6 .
⇒ P = C 3!
3 3
.
6 P = C 3! is the required equation relating parts (a) and (b).
6
3 3
.
n
n
In general, we can deduce that P = C r!.
r r
78

02 STPM Math(T) T3.indd 78 28/10/2021 10:21 AM