4.1 Permutations Part 1

(A) rs Multiplication Principle/ Rule

1. If an operation can be carried out in r ways and another operation can be carried out in s ways, then the number of ways to carry out both the operations consecutively is r × s, i.e. rs.

2. The rs multiplication principle can be expanded to three or more operations. If the numbers of ways for the occurrence of events A, B and C are r, s and p respectively, the number of ways for the occurrence of all the three events consecutively is r × s × p, i.e. rsp.

Example 1:
There are 3 different roads to travel from town P to town Q and 4 different roads to travel from town Q to town R. Calculate the number of ways a person can travel from town P to town R via town Q.

Solution:
3 × 4 = 12

(B) Permutations

Example 2:
Calculate each of the following.
(a) 7!
(b) 4!6!
(c) 0!5!
$\begin{array}{l}\left(d\right)\frac{7!}{5!}\\ \left(e\right)\frac{8!}{4!}\\ \left(f\right)\frac{n!}{\left(n-2\right)!}\\ \left(g\right)\frac{n!0!}{\left(n-1\right)!}\\ \left(h\right)\frac{3!\left(n+1\right)!}{2!n!}\end{array}$

Solution:
(a) 7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040
(b) 4!6! = (4 × 3 × 2 × 1)( 6 × 5 × 4 × 3 × 2 × 1) = 17280
(c) 0!5! = (1)( 5 × 4 × 3 × 2 × 1) = 120

$\begin{array}{l}\text{(d)}\frac{7!}{5!}=\frac{7\text{}×6\text{}×5!}{5!}=7×6=42\\ \\ \text{(e)}\frac{8!}{4!}=\frac{8\text{}×7\text{}×6\text{}×5\text{}×4!}{4!}=8×7×6×5=1680\\ \\ \text{(f)}\frac{n!}{\left(n-2\right)!}=\frac{n\left(n-1\right)\left(n-2\right)}{\left(n-2\right)}=n\left(n-1\right)\\ \\ \text{(g)}\frac{n!0!}{\left(n-1\right)!}=\frac{n\left(n-1\right)\left(1\right)}{\left(n-1\right)}=n\\ \\ \text{(h)}\frac{3!\left(n+1\right)!}{2!n!}=\frac{3×2!\left(n+1\right)\left(n\right)\left(n-1\right)}{2!n\left(n-1\right)}=3\left(n+1\right)\end{array}$

Calculator Computation: