EQN-05-07

Permutations vs Combinations

Decide whether order matters, then count selections correctly — the distinction that trips up most counting questions.

~5 min read•Difficulty 3/5

Does the order of the chosen items matter?

Permutations of r from n : n \times (n - 1) \times \dots (r factors)

Combinations of r from n : that permutation count \div r!

The division by r! removes the duplicate orderings of the same group.

Choosing 2 people from 5:

ordered (e . g . president, vice) : 5 \times 4 = 20

unordered (a committee of 2) : 20 \div 2! = 20 \div 2 = 10

"How many handshakes among n people" is choosing 2 with order irrelevant $: n (n - 1) /2.$

6 people \to 6 \times 5/2 = 15 handshakes .

Diagram

Eight people each shaking hands once $= C (8, 2) = 28$ lines.

Worked examples

Example 1 (difficulty 3/5)

How many ways can a committee of 3 be chosen from 6 people?

◎ Predict the answer first:

Example 2 (difficulty 2/5)

At a meeting of 8 people, everyone shakes hands once with everyone else. How many handshakes?

◎ Predict the answer first:

Counting a committee as if order mattered

Why: The stage-by-stage method gives ordered counts.

Fix: Divide by r! when the chosen group is unordered.

Forgetting to halve the handshake count

Why: Counting each person’

s n - 1

handshakes double-counts every pair.

Fix: Divide by

2 : n (n - 1) /2.