Page 7 - Pra U STPM 2022 Penggal 3 - Maths (T)

Mathematics Semester 3 STPM Chapter 2 Probability
.
n P = n(n – 1)(n – 2) … (n – r + 1) (n – r)(n – r – 1) … 2 1
r (n – r)(n – r – 1) … .
2 1
.
n(n – 1)(n – 2) … (n – r + 1)(n – r)(n – r – 1) … 2 1
=
2 1
(n – r)(n – r – 1) … .
= n!
(n – r)!
Second permutation rule
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The number of permutations of n distinct elements taken r at a time is
n!
n P = (n – r)! . 2
r

Example 5

Evaluate each of the following permutations.
6
(a) 8 P 2 (b) 12 P 9 (c) 5 P 5 (d) P 1

Solution: (a) 8 P = 8!
2 (8 – 2)!
= 8!
6!
= 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
6 × 5 × 4 × 3 × 2 × 1
= 8 × 7
= 56
or
8 P = 8 × 7 = 56 by using multiplication principle.
2
(b) P = 12!
12
9 (12 – 9)!
= 12!
3!
= 12 × 11 × 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
= 79 833 600 3 × 2 × 1

5
(c) P = 5!
5 (5 – 5)!
= 5!
0! 0! = 1
= 5!
= 120

(d) P = 6!
6
1 5!
= 6
Note: 1. Example 5(c) demonstrates the first permutation rule, P = n!.
n
n
2. Example 5(d) illustrates the case for selecting one element from a set of n elements, i.e. P = n.
n
1
75

02 STPM Math(T) T3.indd 75 28/10/2021 10:21 AM