Permutation & Combination
Permutation and combination is organised counting. Decide one thing before touching a formula: does order matter? Arrangements are permutations, selections are combinations, and every restriction — together, apart, a fixed place — is dealt with before the free choices.
Updated 11 October 2026 · Shows up in some papers rather than all, usually as one or two arrangement or selection questions.
How to think about permutation & combination
Order matters, or it does not
Choosing a captain and a vice-captain from 11 players is a permutation: A as captain with B as vice-captain is a different outcome from B with A. Choosing any 2 of the 11 for a fielding drill is a combination: {A, B} is one selection however you list it. The test is simple. Swap two of the chosen items; if that gives a new outcome, use permutations, and if not, combinations.
Multiply for "and", add for "or"
When a task happens in stages, multiply: 3 shirts and 4 trousers give 3 × 4 = 12 outfits. When it can happen one way or another, with no overlap, add the ways. Most committee questions are a sum of products, such as (2 men and 1 woman) + (1 man and 2 women), and "at least one" is usually faster as total − none.
Deal with the restriction first
If some items must be together, glue them into one block, arrange the blocks, then multiply by the ways to order the inside of the block. If a place is restricted, such as a first digit that cannot be 0 or vowels only at odd positions, fill that place first and the free places after. For two items that must never be together, take total − together; for "no two of them together" with three or more, place the others first and drop these into the gaps.
Repeated letters and round tables
Identical items cannot be told apart, so divide by the factorial of each repeat count: BALLOON has 7! / (2! × 2!) = 1260 arrangements, because L and O each appear twice. A circle has no first seat, so fix one person and arrange the rest: (n − 1)!. If the arrangement can also be flipped over, like beads on a necklace, halve it.
Formulas and shortcuts
- Factorial
n! = n × (n − 1) × … × 2 × 10! = 1. Knowing up to 7! = 5040 by heart saves time.
- Permutations (order matters)
ⁿPᵣ = n! / (n − r)!¹⁰P₂ = 10 × 9 = 90. Start at n and multiply r numbers going down.
- Combinations (order does not matter)
ⁿCᵣ = n! / (r! × (n − r)!)ⁿCᵣ = ⁿPᵣ / r!, and ⁿCᵣ = ⁿC(n − r): ¹⁵C₁₃ = ¹⁵C₂ = 105. Always work out the smaller side.
- Arrangements with repeated items
n! / (p! × q! × …)p, q, … are the number of times each item repeats.
- Circular arrangements
(n − 1)!Use (n − 1)! / 2 when clockwise and anticlockwise count as the same, as with a necklace or garland of 3 or more beads.
- k items always together
(n − k + 1)! × k!The block counts as one item and k! orders it inside. Total − this gives "not all together".
- At least one from n different items
2ⁿ − 1Each item is either in or out (2ⁿ ways), minus the empty selection.
Solved examples
Q1. In how many ways can the letters of the word LEADING be arranged so that the vowels always come together?
- LEADING has 7 different letters. Vowels: E, A, I. Consonants: L, D, N, G.
- Glue the 3 vowels into one block. That leaves 5 units to arrange: L, D, N, G and [EAI].
- 5 units in a row: 5! = 120 ways
- Inside the block the vowels can be ordered in 3! = 6 ways
- Total = 120 × 6
Answer: 720 ways
With no restriction there are 7! = 5040 arrangements. So the vowels are together in 720 of them, and "vowels not all together" gives 5040 − 720 = 4320.
Q2. A committee of 3 is to be formed from 5 men and 4 women. In how many ways can it be formed if it must include at least one woman?
- Members have no separate roles, so order does not matter: combinations.
- "At least one woman" is three cases. One subtraction is faster: total − committees with no woman.
- All committees: ⁹C₃ = (9 × 8 × 7) / (3 × 2 × 1) = 84
- Committees with no woman (all men): ⁵C₃ = 10
- 84 − 10
Answer: 74 ways
Case by case: 1 woman and 2 men = 4 × 10 = 40; 2 women and 1 man = 6 × 5 = 30; 3 women = 4. Then 40 + 30 + 4 = 74.
Q3. How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 if no digit is repeated? How many of these are even?
- Order matters (123 and 321 are different numbers), so this is a permutation.
- All numbers: ⁵P₃ = 5 × 4 × 3 = 60
- For even numbers, fill the restricted place first. The units digit must be 2 or 4: 2 ways.
- Then the hundreds digit has 4 choices left and the tens digit 3: 2 × 4 × 3 = 24
Answer: 60 numbers in all, of which 24 are even
Cross-check: 2 of the 5 digits are even, and each digit lands in the units place equally often, so 2/5 × 60 = 24.
Practice questions with answers
1. In how many ways can a president and a vice-president be chosen from a club of 12 members?
Answer: C. 132 — The two posts are different, so order matters: ¹²P₂ = 12 × 11 = 132. ¹²C₂ = 66 only picks the pair, not who gets which post.
2. In how many different ways can the letters of the word BANANA be arranged?
Answer: A. 60 — BANANA has 6 letters with A three times and N twice: 6! / (3! × 2!) = 720 / 12 = 60.
3. In how many ways can 7 people be seated around a round table?
Answer: B. 720 — Fix one person to remove the rotations, then arrange the other 6: (7 − 1)! = 6! = 720. Halving to 360 is only for arrangements that can be flipped over, such as beads on a necklace.
4. A committee of 5 is to be chosen from 6 men and 4 women so that it has exactly 2 women. In how many ways can this be done?
Answer: B. 120 — Choose the women and the men separately, then multiply: ⁴C₂ × ⁶C₃ = 6 × 20 = 120. Adding instead, 6 + 20 = 26, is wrong because both choices are part of the same committee.
5. How many 3-digit numbers can be formed from the digits 0, 1, 2, 3 and 4 if no digit is repeated?
Answer: A. 48 — The hundreds place cannot be 0, so fill it first: 4 ways. Tens place: 4 digits left, since 0 is now allowed. Units place: 3 left. 4 × 4 × 3 = 48. ⁵P₃ = 60 also counts strings like 012, which are not 3-digit numbers.
6. In how many ways can 4 boys and 3 girls stand in a row so that no two girls are next to each other?
Answer: C. 1440 — Arrange the boys first: 4! = 24. They leave 5 gaps (_ B _ B _ B _ B _). Put the 3 girls into 3 different gaps, in order: ⁵P₃ = 60. Total = 24 × 60 = 1440. 4320 answers "not all three together", which is a different condition.
7. At a meeting, everyone shakes hands with everyone else exactly once. There are 66 handshakes in all. How many people are at the meeting?
Answer: D. 12 — Each handshake is one pair of people, so handshakes = ⁿC₂ = n(n − 1)/2. Then n(n − 1) = 132 = 12 × 11, so n = 12.
8. In how many ways can the letters of the word MOTHER be arranged so that the vowels occupy only the odd positions?
Answer: D. 144 — The vowels O and E must go into 2 of the 3 odd places (1st, 3rd, 5th), in order: ³P₂ = 6. The 4 consonants fill the other 4 places: 4! = 24. Total = 6 × 24 = 144.
Where marks get lost
- Using combinations when the chosen items get different roles, such as captain and vice-captain or first and second prize.
- Adding the counts for choices that happen together; the men and the women chosen for one committee multiply.
- Letting 0 take the first place in a digit question, which counts 2-digit numbers as 3-digit ones.
- Forgetting to order the inside of a glued block (× k!) or to divide out repeated letters (÷ p!).
Reading is not practising.
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