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# Statistics I · Week 6 — Permutations and combinations Deep study for Quiz 2 week 6. **Permutations** count ordered arrangements; **combinations** count unordered selections.

Statistics I · Week 6 — Permutations and combinations

Deep study for Quiz 2 week 6. Permutations count ordered arrangements; combinations count unordered selections.

Week map

Order matters? → permutation → order irrelevant? → combination → formulas → complement counting → word-problem translation.

Permutation notation

$P(n, r)$ or ${}^nP_r$ → permutations of $r$ items chosen from $n$ distinct → order matters.
Formula: $P(n,r) = \dfrac{n!}{(n-r)!}$ .
Special case: $P(n,n) = n!$ — arrange all $n$ .

Mini-example: Top 2 finishers from 5 runners. Order matters (gold vs silver):

P(5,2) = 5 \times 4 = 20

Combination notation

$C(n, r)$ or $\binom{n}{r}$ → combinations — choose $r$ from $n$ → order does not matter.
Formula: $C(n,r) = \dfrac{n!}{r!(n-r)!}$ .
$\binom{n}{0} = 1$ ; $\binom{n}{n} = 1$ .

Mini-example: Choose 2 toppings from 5 regardless of order:

C(5,2) = \frac{5 \times 4}{2 \times 1} = 10

Decision rule

Question phrasing	Use
Arrange, rank, sequence, PIN order	Permutation
Choose, select, committee, hand	Combination

Pattern families

Easy — Small n compute directly

List or multiply descending factors:

5 \times 4 \times 3

for

P(5,3)

. Divide by

r!

when order ignored.

Medium — Formula application

Plug into

P(n,r)

C(n,r)

. Simplify factorials before multiplying large numbers.

Hard — Complement or restrictions

“At least one” → total minus “none.” Adjacent or fixed-position constraints may reduce to smaller permutation subproblem.

Worked mini-examples

Example 1 — Permutation.

P(4,2) = 4 \times 3 = 12

Example 2 — Combination.

\binom{6}{2} = \frac{6 \times 5}{2} = 15

Example 3 — Compare.

3-letter code from {A,B,C,D} without repeat:

P(4,3) = 24

. Choose 3 letters unordered:

\binom{4}{3} = 4

Example 4 — Committee.

Choose 3 from 7 people for roles-free committee:

\binom{7}{3} = 35

Example 5 — Complement.

Choose at least one red from 3 red and 4 blue (choose 2 total): total

\binom{7}{2}=21

minus no red

\binom{4}{2}=6

→ 15.

Traps

Using combination when order matters (rankings, passwords).
Using permutation when order irrelevant (teams, subsets).
Forgetting to divide by $r!$ when converting ordered to unordered.
$C(n,r) = C(n, n-r)$ symmetry — useful check.
Identical items need division beyond basic $C(n,r)$ (not all distinct).

Diagnostic (try yourself)

How many ways to arrange 3 distinct books on a shelf?
How many 2-person committees from 6 people (order irrelevant)?
Compute $P(5,3)$ and $C(5,3)$ . Which is larger?
Word problem: “President and VP from 8 candidates.” Permutation or combination? Count?
$\binom{10}{0}$ equals what?