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Permutation vs Combination: When Order Matters, Repeated Letters and Restrictions

Counting principle • nPr versus nCr • Words with repeated letters • Together and never together • Circular arrangements • Committees and 'at least one'

SSC Banking 📅 Last updated: 2026-10-23

Every counting question rests on one rule. If one choice can be made in m ways and a second, independent choice in n ways, the two together can be made in m × n ways. Permutations and combinations are this rule applied to choices that use up items as you go.

Worked example 1: outfits

With 3 shirts and 4 pairs of trousers, there are 3 × 4 = 12 outfits.

Worked example 2: books on a shelf

5 different books can be arranged in 5 × 4 × 3 × 2 × 1 = 5! = 120 ways: 5 choices for the first place, 4 for the next, and so on.

The only question to ask is whether order matters. If swapping two chosen items gives a different result, count permutations. If it does not, count combinations.

Permutation (arrangement)Combination (selection)
Order matters?YesNo
FormulanPr = n! ÷ (n − r)!nCr = n! ÷ (r! (n − r)!)
Typical wordingarrange, seat, rank, posts, passwordschoose, select, team, committee

Each selection of r items can be arranged in r! orders, so nPr = nCr × r!.

Worked example 3: a committee

A committee of 3 from 7 people: order does not matter, so 7C3 = 35.

A captain and a vice-captain from the same 7: the two posts differ, so order matters and 7P2 = 7 × 6 = 42.

Number-formation questions are permutations with a twist: you must check whether digits may repeat. Without repetition, each place has one choice fewer than the last; with repetition, every place has the full set of choices.

Worked example 4: three-digit numbers from 1 to 5

Without repeating a digit: 5 × 4 × 3 = 60 numbers.

If digits may repeat, every place has 5 choices: 5³ = 125 numbers. If 0 is among the digits, remember that it cannot stand in the first place.

To count arrangements of the letters of a word, start with n! for n letters, then divide by p! for every letter that appears p times, because swapping identical letters does not create a new word.

Worked example 5: LEVEL

LEVEL has 5 letters, with L twice and E twice. Arrangements = 5! ÷ (2! × 2!) = 120 ÷ 4 = 30.

When some items must stay together, tie them into a single block. Arrange the block with the other items, then multiply by the number of ways to arrange the items inside the block.

Worked example 6: vowels together in MOTHER

The vowels O and E form one block. That leaves 5 units (the block plus M, T, H, R), arranged in 5! = 120 ways, and the block itself in 2! ways: 120 × 2 = 240.

For "never together", subtract the together count from the total. It is usually easier than counting the separated arrangements directly.

Around a circle there is no first seat: rotating everyone one place gives the same arrangement. Fix one person's seat and arrange the others in a line, giving (n − 1)! arrangements. If clockwise and anticlockwise orders count as the same, as with beads on a necklace, divide by 2 as well.

Worked example 7: six at a round table

6 people can sit around a round table in (6 − 1)! = 5! = 120 ways.

When a committee must contain a set number from each group, choose from each group separately and multiply.

Worked example 8: 3 men and 2 women

From 5 men and 4 women, a committee of 3 men and 2 women can be formed in 5C3 × 4C2 = 10 × 6 = 60 ways.

"At least one" questions are fastest by complement: count all selections, then subtract those with none of the required kind.

Worked example 9: at least one woman

A committee of 3 from 4 men and 3 women: all selections 7C3 = 35, with no women 4C3 = 4. At least one woman: 35 − 4 = 31.

  • ✘ Using nPr for a committee or team, where order does not matter.
  • ✘ Forgetting to arrange the items inside a "together" block.
  • ✘ Using n! instead of (n − 1)! for seats around a circle.
  • ✘ Adding the cases for "at least one" one by one when the complement is quicker.

Counting is the first step in probability with coins, dice and cards, and in circular seating arrangements. The formula list is in the Quant cheat code.

Q1. In how many ways can 3 students be chosen from a group of 8? easy

✅ Answer: (c) 56

💡 Explanation: Choosing a group ignores order, so use 8C3 = (8 × 7 × 6) ÷ (3 × 2 × 1) = 56. 336 is 8P3, which would count the same three students in different orders.

Q2. In how many ways can gold, silver and bronze medals be awarded among 6 runners? easy

✅ Answer: (a) 120

💡 Explanation: The medals are different, so order matters: 6 choices for gold, then 5 for silver, then 4 for bronze. 6P3 = 6 × 5 × 4 = 120.

Q3. How many different arrangements can be made of the letters of the word BANANA? medium

✅ Answer: (d) 60

💡 Explanation: BANANA has 6 letters with A repeated 3 times and N twice. Arrangements = 6! ÷ (3! × 2!) = 720 ÷ 12 = 60.

Q4. In how many ways can 7 people sit around a circular table? medium

✅ Answer: (b) 720

💡 Explanation: Around a circle, rotations of the same seating are identical, so fix one person and arrange the rest: (7 − 1)! = 6! = 720.

Q5. In how many arrangements of the letters of PENCIL do the vowels never come together? hard

✅ Answer: (a) 480

💡 Explanation: All arrangements: 6! = 720. With E and I together as one block: 5! × 2! = 240. Never together = 720 − 240 = 480.

Q6. A committee of 4 is chosen from 6 men and 4 women. In how many ways can it have exactly 2 women? medium

✅ Answer: (c) 90

💡 Explanation: Choose the 2 women and the 2 men independently and multiply: 4C2 × 6C2 = 6 × 15 = 90.

How do I decide quickly between permutation and combination?

Ask whether swapping two chosen items gives a different outcome. If it does, as with ranks, seats, passwords or different posts, order matters and you need a permutation. If it does not, as with teams, committees or handfuls, use a combination.

Why do I divide by the factorials of repeated letters?

Swapping two identical letters gives the same word. If a letter appears p times, each distinct word has been counted p! times among the n! arrangements, so you divide by p! for each repeated letter.

All problems and solutions on this page are original. No outside facts are relied on; every numeric answer is recomputed by software before publishing.

How this page was prepared. Drafted with AI assistance (Claude (Anthropic AI model)) from the sources listed above. 17 calculations on this page were recomputed by software and matched. Automated checks also covered structure, internal links and duplicate text. Software checks cannot confirm facts. Reviewed by praveenkumar on 2026-10-23 against the sources listed above.

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