IMAT Mathematics · Probability and Statistics

Counting: Arrangements and Choices

Count outcomes without listing them: multiply the number of ways for each step, divide out repeats, and use combinations when order does not matter.

All Probability and Statistics notes

Why this matters

A 2017 question asked how many ways the letters of a short word with repeated letters can be arranged. Counting also gives the totals you divide by in every probability question.

Concept 1 of 2: The multiplication principle and arrangements (permutations)

If you choose a shirt from 4 and then trousers from 3, each shirt goes with each pair of trousers:

4 \times 3 = 12

outfits. Arranging objects in a row is the same idea: the first place has

n

choices, the next

n - 1

, and so on. When some objects are identical, swapping them makes no new arrangement, so you divide those swaps out.

Definition

Multiplication principle: if one step can be done in $m$ ways and the next in $n$ ways, the two together can be done in $m \times n$ ways.
$n$ factorial: $n! = n \times (n - 1) \times \cdots \times 2 \times 1$ , with $0! = 1$ . It counts the arrangements of $n$ different objects in a row.
Permutations of $r$ objects chosen from $n$ different ones, in order: ${}^nP_r = \dfrac{n!}{(n - r)!}$ .
Repeated objects: $n$ objects with $p$ alike of one kind, $q$ alike of another, and so on, can be arranged in $\dfrac{n!}{p!\,q!\cdots}$ ways.

Arrangements

{}^nP_r = \frac{n!}{(n - r)!} \qquad \text{with repeats: } \frac{n!}{p!\,q!\cdots}

$n$ number of objects available
$r$ number placed in order
$p, q$ sizes of groups of identical objects

Worked example

In how many ways can the letters of the word PEPPER be arranged? How many of these arrangements begin with R?

Practice this conceptself-check · 4 quick reps

Identical letters do not make new arrangements

Swapping two identical letters gives the same word, so

n!

overcounts. Divide by the factorial of the size of each group of repeats; forgetting this gives a much larger number that IMAT puts among the options.

Concept 2 of 2: Combinations: choosing a group when order does not matter

Choosing a team of 3 from 8 people is different from filling 3 named posts: the team Ana, Ben, Carla is the same team in any order. Count the ordered choices and divide by the

3!

orders of each team.

Definition

A combination is a selection where order does not matter.
The number of ways to choose $r$ from $n$ different objects is $\dbinom{n}{r} = {}^nC_r = \dfrac{n!}{r!\,(n - r)!}$ .
$\dbinom{n}{r} = \dbinom{n}{n - r}$ : choosing who is in is the same as choosing who is out.
For a group from two separate pools, multiply: choose from each pool, then use the multiplication principle.

Combinations

\binom{n}{r} = \frac{n!}{r!\,(n - r)!}

$n$ number of objects to choose from
$r$ number chosen

Worked example

A committee of 4 is to be formed from 5 women and 4 men, with exactly 2 women. In how many ways? And how many committees of 3 can be formed from 8 people with no condition?

Practice this conceptself-check · 4 quick reps

Ask whether order matters before you count

Named posts, rankings, codes and words are ordered: use permutations. Teams, hands of cards and sets of toppings are not: use combinations. The two answers differ by a factor of

r!

, and both appear among the options.

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

The multiplication principle and arrangements (permutations)
Arrangements
${}^nP_r = \frac{n!}{(n - r)!} \qquad \text{with repeats: } \frac{n!}{p!\,q!\cdots}$
Combinations: choosing a group when order does not matter
Combinations
$\binom{n}{r} = \frac{n!}{r!\,(n - r)!}$

Watch out for (2)

Identical letters do not make new arrangements
→ The multiplication principle and arrangements (permutations)
Ask whether order matters before you count
→ Combinations: choosing a group when order does not matter

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