# Permutations for Counting Techniques

\large \require{AMSsymbols} \displaystyle \begin{align} ^n P_r &= \frac{n!}{(n-r)!} \\ &= \overbrace{n \times (n-1) \times (n-2) \times \cdots }^{r} \end{align}

## Example

$^{10} P _3 = \overbrace{10 \times 9 \times 8}^{3} = 720$

$\displaystyle ^{10} P _3 = \frac{10!}{(10-3)!} = \frac{10 \times 9 \times 8 \times 7!}{7!} = 10 \times 9 \times 8 = 720$

## Question 1

Evaluate the following.

(a)     $^9 P _2$

$=9 \times 8 = 72$

(b)     $^5 P _1$

$=5$

(c)     $^6 P _0$

$\displaystyle =\frac{6!}{(6-0)!} = 1$

(d)     $^4 P _4$

$=4 \times 3 \times 2 \times 1 = 4! = 24$

(e)     $^5 P _6$

$\text{undefined}$

(f)     $^5 P _{-1}$

$\text{undefined}$

## Question 2

Find the value of $n$, if $^6 P_n = 120$.

\begin{align} 6 \times 5 &= 30 \\ 6 \times 5 \times 4 &= 120 \\ ^6 P_3 &= 120 \\ \therefore n &=3 \end{align}

## Question 3

The digits $1, \ 2, \ 3$ and $4$ are each written on a card and then placed in a box.

(a)     A card is chosen, read and replaced. A second card is chosen and read. How can many possible two digit-numbers be chosen?

$4 \times 4 = 15$

(b)     A card is chosen, read and not-replaced. A second card is chosen and read. How can many possible two digit-numbers be chosen?

$^4 P_2 = 4 \times 3 = 12$

## Question 4

How many ways are there of choosing three books from a shelf of ten and reading them in order?

$^{10} P_3 = 720$ ways

## Question 5

In how many ways can the letters of the word TODAY be arranged if they are used once only and taken:

(a)     three at a time?

$^5 P_3 = 60$

(b)     four at a time?

$^5 P_4 = 120$

(c)     five at a time?

$^5 P_5 = 120$