Permutations — Arrangements

Section 1 of 3

Arranging in Order

Filling boxes

To count arrangements, draw a box for each position and write how many choices fit each — then multiply (an AND). Each choice used up leaves one fewer.

Worked example — five horses

In how many orders can $5$ horses finish a race (no dead heats)?

$5$ choices for $1$st, then $4$, then $3$, then $2$, then $1$

$5 \times 4 \times 3 \times 2 \times 1 = 120$

$120$

Factorial

$5 \times 4 \times 3 \times 2 \times 1$ is written $5!$ (‘five factorial’). It counts the ways to arrange all of $5$ different things.

Worked example — dogs and girls

(i) In how many orders can $6$ dogs line up? (ii) In how many orders can $10$ girls line up?

(i) $6! = 6\times5\times4\times3\times2\times1 = 720$

(ii) $10! = 3{,}628{,}800$

$720;\ 3{,}628{,}800$

Section 2 of 3

Choosing Some, Not All

AND / OR still apply

AND = multiply, OR = add. When you only fill some of the positions, just multiply the choices for the boxes you fill.

Worked example — three-letter words

How many $3$-letter words can be made from $A, B, C, D, E$, each letter used once?

$3$ boxes: $5$ choices, then $4$, then $3$

$5 \times 4 \times 3 = 60$

$60$

Section 3 of 3

Numbers

Worked example — two-digit numbers

Using the digits $3, 4, 5, 6, 7, 8$ (each once), how many $2$-digit numbers can be made? How many are odd?

All: $6 \times 5 = 30$

Odd ⇒ last digit is odd ($3, 5, 7$): $5 \times 3 = 15$ (fill the last box first: $3$ odd choices, then $5$ for the first)

$30;\ 15$

You try

Using the digits $4, 5, 6, 7, 8$ (each once), find how many $2$-digit numbers (i) in total; (ii) start with $8$; (iii) are odd.

For ‘odd’ fill the last box first (odd digits: $5, 7$).

(i) $5 \times 4 = 20$

(ii) $1 \times 4 = 4$

(iii) last box odd ($5$ or $7$): $4 \times 2 = 8$

$20;\ 4;\ 8$

That’s Class 11.

Filling boxes, factorials ($n!$), and arranging some of a set. Class 12: conditions and keeping items together.