4.9 Permutations & Combinations

The Two Counting Principles

Multiplication principle. If a first task can be done in $m$ ways and, for each of those, a second in $n$ ways, the pair can be done in $m \times n$ ways. Use it when tasks happen in sequence — signalled by the word and.

Addition principle. If a task can be done in $m$ ways or in $n$ ways, with no overlap, the total is $m + n$. Use it for mutually exclusive alternatives — signalled by the word or.

Factorials

$$n! = n \times (n-1) \times (n-2) \times \cdots \times 2 \times 1, \qquad 0! = 1$$

The convention $0! = 1$ is not arbitrary; it is what makes the formulae below work at their boundaries, and it appears as an objective item in its own right.

Permutations: Order Matters

The number of ways of arranging $r$ objects chosen from $n$ distinct objects is

$$^{n}P_{r} = \frac{n!}{(n-r)!}$$

Example. In how many ways can a first, second and third prize be awarded among 6 candidates? Order matters because the prizes differ:

$$^{6}P_{3} = \frac{6!}{3!} = 6\times5\times4 = 120$$

Combinations: Order Does Not Matter

The number of ways of selecting $r$ objects from $n$ is

$$^{n}C_{r} = \frac{n!}{r!\,(n-r)!} = \frac{^{n}P_{r}}{r!}$$

Example. In how many ways can a committee of 3 be chosen from 6 people?

$$^{6}C_{3} = \frac{6\times5\times4}{3\times2\times1} = 20$$

Note that $120 = 20 \times 3!$ — each selection of three can be arranged in $3! = 6$ orders, which is precisely the difference between the two counts.

Properties of combinations

$$^{n}C_{r} = {}^{n}C_{n-r}, \qquad ^{n}C_{0} = {}^{n}C_{n} = 1, \qquad ^{n}C_{r} + {}^{n}C_{r-1} = {}^{n+1}C_{r}$$

The first identity is a genuine time-saver: computing $^{20}C_{18}$ as $^{20}C_{2} = 190$ takes seconds.

The Deciding Question

A committee is a combination; a committee with designated posts such as chairman and secretary is a permutation, or a combination followed by an arrangement.

Arrangements with Repetition

When some objects are identical, divide by the factorial of each repetition count:

$$\text{Arrangements} = \frac{n!}{p!\,q!\,r!\cdots}$$

Example. How many distinct arrangements of the letters of ENGINEER?

The word has 8 letters with E appearing 3 times and N appearing 2 times:

$$\frac{8!}{3!\,2!} = \frac{40320}{12} = 3360$$

Circular Arrangements

Seating $n$ distinct people around a round table gives

$$(n-1)!$$

arrangements, because rotating everyone by one seat does not create a new arrangement — only relative order matters. Fix one person's position and arrange the remaining $n-1$.

If clockwise and anticlockwise arrangements are not distinguished, as with a garland of flowers or a necklace of beads, halve the result:

$$\frac{(n-1)!}{2}$$

Example. Six engineers sit around a circular table: $(6-1)! = 120$ ways. If instead six identical-looking beads form a necklace: $5!/2 = 60$.

Restricted Arrangements

Objects that must stay together

Treat the block as a single unit, arrange the units, then arrange within the block.

Example. In how many ways can 5 people be arranged in a row if two particular people must sit together? Treat the pair as one unit, giving 4 units arranged in $4! = 24$ ways, and the pair internally in $2! = 2$ ways. Total $= 48$.

Objects that must be separated

Count the total and subtract the together case.

Example. With the same 5 people, the number of arrangements in which the two are not together is $5! - 48 = 120 - 48 = 72$.

Fixed positions

Example. How many 4-digit even numbers can be formed from the digits 1 to 7 without repetition? The units digit must be 2, 4 or 6, giving 3 choices, and the remaining three positions are filled from the other 6 digits in $^{6}P_{3} = 120$ ways. Total $= 3 \times 120 = 360$.

Selection Problems

Example. From 7 men and 5 women, how many committees of 5 can be formed with exactly 3 men?

$$^{7}C_{3}\times{}^{5}C_{2} = 35\times10 = 350$$

Example. How many committees of 5 contain at least 3 men? Sum the cases of exactly 3, 4 and 5 men:

$$^{7}C_{3}{}^{5}C_{2} + {}^{7}C_{4}{}^{5}C_{1} + {}^{7}C_{5}{}^{5}C_{0} = 350 + 175 + 21 = 546$$

The phrase at least always means summing cases, or subtracting the complement from the total when that is shorter.

Standard Results Worth Recall

All of these appear regularly, and each takes a few seconds once recognised.