Permutation vs combination — when to use which with examples

Question

In how many ways can 3 students be chosen from a group of 8 to form a committee? In how many ways can a president, secretary, and treasurer be chosen from the same group?

(CBSE 11 / JEE Main — Permutations and Combinations)

P vs C Decision Tree

flowchart TD
    A["Counting Problem"] --> B{"Does order matter?"}
    B -->|"Yes — arrangement matters"| C["PERMUTATION"]
    B -->|"No — only selection matters"| D["COMBINATION"]
    C --> C1["nPr = n! / (n-r)!"]
    D --> D1["nCr = n! / r!(n-r)!"]
    C --> E["Examples: ranking, seating, passwords"]
    D --> F["Examples: committees, teams, choosing items"]
    A --> G{"Repetition allowed?"}
    G -->|Yes| H["n^r (permutation) or n+r-1 C r (combination)"]
    G -->|No| C
    G -->|No| D

Solution — Step by Step

Choosing 3 students from 8 for a committee — the order does not matter. Choosing {A, B, C} is the same as choosing {C, A, B}.

This is a combination problem:

\binom{8}{3} = \frac{8!}{3! \times 5!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = \frac{336}{6} = \mathbf{56}

Choosing 3 people for 3 specific roles — the order matters. Making A president and B secretary is different from making B president and A secretary.

This is a permutation problem:

P(8, 3) = \frac{8!}{(8-3)!} = \frac{8!}{5!} = 8 \times 7 \times 6 = \mathbf{336}

P(n, r) = \binom{n}{r} \times r!

336 = 56 \times 6 = 56 \times 3!

The permutation count is always $r!$ times the combination count, because each combination of $r$ items can be arranged in $r!$ ways.

Why This Works

Combinations count the number of ways to select items, ignoring the order. Permutations count the number of ways to arrange items, where order matters. The factorial in the denominator of the combination formula ( $r!$ ) divides out the duplicate arrangements.

The one question to ask yourself: “If I rearrange the selected items, does it count as a different outcome?” If yes, use permutation. If no, use combination.

Alternative Method — Slot Filling

For permutations, think of filling slots:

President: 8 choices
Secretary: 7 remaining choices
Treasurer: 6 remaining choices
Total: $8 \times 7 \times 6 = 336$

This “multiply the choices” approach is often more intuitive than the formula.

In JEE, many counting problems combine both P and C. Example: “Choose a committee of 5 from 10, then arrange them in a line.” First, $\binom{10}{5}$ (combination for selection), then $5!$ (permutation for arrangement). Total = $\binom{10}{5} \times 5!$ = $P(10, 5)$ . Recognising which step needs P and which needs C is the real skill.

Common Mistake

The most common error: using permutation when the problem asks for a committee or group (where order does not matter). The word “committee” or “team” almost always signals combination. Words like “arrange,” “rank,” “password,” or “code” signal permutation. Train yourself to look for these keywords — they tell you the formula before you even set up the problem.