Permutation and Combination Basics

Overview

Permutation and Combination Basics appears about 0.5 times per paper across the last four AFCAT solved papers, placing it in the solid add-on band of Numerical Ability.

Permutation and Combination tests whether you can read a counting problem, decide if order matters, and then apply the correct factorial expression. In AFCAT the topic appears in roughly one paper out of two and contributes around half a mark per paper on average. Across the past several cycles the questions have stayed at the basic counting level — selecting a committee, arranging letters of a word, seating people in a row or around a table, forming numbers from digits. Advanced combinatorics such as derangements, integer partitions, or inclusion-exclusion on three sets does not appear.

The single most important habit to build is the order check. Before you pick a formula, ask one question — does swapping two chosen items give a new arrangement or the same selection? If swapping changes the outcome, you need a permutation. If swapping gives the same group, you need a combination. Get that decision right and the rest is one factorial division. Get it wrong and you will either be off by a factor of r! or end up with one of the distractor answers the examiner planted exactly for this mistake.

Why P&C tests structural thinking, not just formulas

Most numerical topics reward arithmetic speed. P&C is different — the arithmetic is tiny but the setup is everything. The examiner can write almost any P&C problem in two ways that look identical but need different formulas. The difference is hidden in one word, usually arrange versus select, or seat in a row versus seat around a circular table.

• Order matters → permutation. Arranging books on a shelf, seating people, forming a number from digits, electing president and vice-president.
• Order does not matter → combination. Forming a committee, choosing a subset of fruits, picking a team of eleven from fifteen.
• Repetition allowed → multiplication principle. Choosing a 4-digit PIN with digits 0–9 repeatable gives 10⁴.
• Repetition of identical objects → divide by factorial. Arranging the letters of BOOK is 4!/2! because the two Os are interchangeable.

Permutations vs Combinations — when each applies

The cleanest way to remember the difference is to test with a concrete tiny example. Take three objects A, B, C and ask for groups of two.

• Permutations of 2 from {A, B, C}: AB, BA, AC, CA, BC, CB — that is 6 arrangements. Formula: 3P2 = 3!/1! = 6.
• Combinations of 2 from {A, B, C}: {A,B}, {A,C}, {B,C} — that is 3 selections. Formula: 3C2 = 3!/(2!1!) = 3.

Notice that nPr is always r! times larger than nCr. That is because each combination can be internally rearranged in r! ways, and a permutation counts every such rearrangement as a separate outcome. The relationship is exact:

nPr = nCr × r!

This identity is useful as a sanity check. If you computed a permutation answer and the choices include another value that is exactly r! times smaller, the smaller one is the combination answer for the same question — useful when you suspect you misread arrange as select.

Factorial — definition and common values

Factorial of a positive integer n, written n!, is the product of all positive integers from 1 to n. By convention 0! is defined as 1; this keeps the combination formula nC0 = 1 consistent. Memorising the small factorials saves real time in the exam hall.

The right working habit is to cancel before multiplying. Take 8! / 6!. Do not expand both. Write it as (8 × 7 × 6!) / 6! = 8 × 7 = 56. Almost every P&C calculation reduces to a few small products once you cancel matching factorials.

Permutation formulas you will actually use

Three permutation formulas cover every AFCAT case.

• Arrangement of all n distinct objects in a row: n!
• Arrangement of r out of n distinct objects (r < n): nPr = n! / (n − r)!
• Arrangement of n objects where some are identical (p of one kind, q of another, r of a third): n! / (p! × q! × r!)

Worked illustrations of nPr for small values are listed below. You should be able to compute these in the head after one read.

The quick rule for nPr by hand: start at n and multiply r consecutive descending integers. nPr therefore has exactly r factors.

Combination formulas and identities

Combinations look harder because of the double factorial in the denominator, but the symmetry identities cut almost every calculation in half.

• nCr = n! / (r! × (n − r)!)
• nC0 = nCn = 1. Choosing nothing or choosing everything is exactly one way.
• nC1 = n. One out of n can be any of the n items.
• nCr = nC(n − r). Choosing r to take is the same as choosing n − r to leave behind. Use this to convert 10C8 into 10C2 = 45, instead of computing 10C8 directly.
• nCr + nC(r − 1) = (n+1)Cr. Pascal's identity — rarely tested directly but it explains the triangle.
• Sum of a row of Pascal's triangle: nC0 + nC1 + nC2 + ... + nCn = 2ⁿ. Useful for "any number of items can be chosen" problems.

If you see 12C10 in a question, write it as 12C2 = 66 before reaching for the calculator method. The symmetry identity is the single highest-return formula in the chapter.

Permutations with repetition — divide by factorial of each repeat

When you arrange the letters of a word that has repeated letters, treating the repeats as distinct overcounts. Two identical letters can swap without changing the visible word, so you must divide the raw n! by the factorial of every repeated-letter count.

Arrangements of n objects with p of one kind, q of another, r of a third = n! / (p! × q! × r!)

The classic AFCAT-style examples:

The arithmetic for MISSISSIPPI: 11! = 11 × 10 × 9 × 8 × 7 × 6 × 5! = 11 × 10 × 9 × 8 × 7 × 6 × 120. Denominator is 4! × 4! × 2! = 24 × 24 × 2 = 1,152. Cancel step by step; the final answer is 34,650. You will not be asked to compute MISSISSIPPI in the exam, but the smaller cousins (BANANA, LEVEL, STATISTICS) do appear.

Permutations with restrictions — gap method and glue method

Restriction problems describe a row arrangement with a side condition — two people must sit together, or never sit next to each other, or a specific seat is fixed. Two clean techniques solve nearly every variant.

Glue method (always together). If two or more items must be adjacent, tie them into a single unit. Count the arrangements of the units, then multiply by the internal arrangements of the tied items.

• Example: 5 people in a row, 2 specified people must sit together. Treat the pair as one block, giving 4 units. Arrangements of 4 units = 4! = 24. The pair can be internally arranged in 2! = 2 ways. Total = 24 × 2 = 48.

Gap method (never together). First arrange the unrestricted items. Then place the restricted items into the gaps between and around them, choosing the required number of gaps.

• Example: 5 boys and 3 girls in a row, no two girls together. Arrange the 5 boys first: 5! = 120 ways. The boys create 6 gaps (one before each boy and one at the end). Place 3 girls in 3 of these 6 gaps with order: 6P3 = 6 × 5 × 4 = 120. Total = 120 × 120 = 14,400.

Fixed position. Slot the fixed item into its forced seat first, then arrange the remaining items normally.

• Example: arrange the letters of EXAM such that E is always first. Lock E. Arrange X, A, M in the remaining 3 positions: 3! = 6 ways.

Circular permutations — (n−1)! and the necklace adjustment

When n distinct people sit around a round table, rotating the entire arrangement by one seat gives the same relative ordering. The exam treats rotations as identical. To compensate, fix one person's seat and arrange the remaining n − 1 in the other seats. The result is (n − 1)! arrangements.

Circular arrangements of n distinct objects = (n − 1)!

If the arrangement is also indistinguishable from its mirror image — typical for a necklace or a garland where the bracelet can be flipped over — divide by 2 once more.

Necklace or garland arrangements of n distinct beads = (n − 1)! / 2

If two specified people must sit together at a round table, glue them as before. The unit count drops by one, so the arrangements become (n − 2)! × 2! after the rotation fix.

Selection problems — committees with at least or with exact constraints

Committee questions are pure combination problems with a side filter. Two sub-types are common.

Exact composition. "Form a committee of 5 from 6 men and 4 women with exactly 3 men." Choose the 3 men and the 2 women separately, then multiply.

• Men: 6C3 = 20. Women: 4C2 = 6. Total = 20 × 6 = 120.

At-least-one constraint. "Form a committee of 4 from 6 men and 4 women with at least 1 woman." The cleaner route is the complement.

• Total committees of 4 from 10 people = 10C4 = 210.
• Committees with no woman = 6C4 = 15.
• Committees with at least one woman = 210 − 15 = 195.

The complement trick saves the trouble of summing C(women, 1) + C(women, 2) + ... case by case. Use it whenever the stem says "at least one" of some category.

Both constraints together. "Committee of 5 from 6 men and 5 women with at least 2 men and at least 2 women." The only valid splits are 2M+3W and 3M+2W.

• 2M, 3W: 6C2 × 5C3 = 15 × 10 = 150.
• 3M, 2W: 6C3 × 5C2 = 20 × 10 = 200.
• Total = 350.

Distributions — putting objects into bins

Distribution problems are rare in AFCAT, but the basic two-case logic is worth knowing.

• Distinct objects into distinct bins, repetition allowed: n^r. Example: 3 letters into 5 letter-boxes — each letter has 5 choices independently, so 5³ = 125 ways.
• Distinct objects into distinct bins, no bin gets more than one: nPr if r ≤ n.
• Identical objects into distinct bins: uses stars-and-bars, but AFCAT does not test this.

If you see a distribution problem in the exam, it will almost certainly be the n^r version. Read carefully — "each letter goes into a different box" signals nPr, while plain "letters into boxes" means repetition is allowed.

Common AFCAT trap patterns to spot in the stem

The examiner has a short list of standard traps. Each one has a tell in the wording.

• Permutation labelled as combination. "In how many ways can 3 prizes — first, second, third — be given to 8 students?" The prizes are distinct, so this is 8P3 = 336, not 8C3 = 56. The distractor 56 will be on the option list.
• Combination labelled as permutation. "In how many ways can a 3-member team be selected from 8 players?" This is 8C3 = 56, not 8P3 = 336. Look for the word team, committee, group, or selection.
• Forgotten repetition. Arranging the letters of LEVEL is 5!/(2!2!) = 30, not 5! = 120. If the word repeats a letter and you wrote a single n!, you have already missed.
• Counting circular as linear. 6 people around a round table is (6−1)! = 120, not 6! = 720. The factor of n separates them.
• Double-counting committees. In a "3 men and 2 women from 5 men and 4 women" problem, do not also multiply by 5! to arrange the committee — the committee is unordered.
• At-least-one without complement. Summing case-by-case is slower and error-prone. Total minus none is the reliable shortcut.
• Leading-zero numbers. "How many 4-digit numbers from digits 0–9 without repetition" — the leading digit cannot be 0. Count first place separately: 9 × 9 × 8 × 7 = 4,536, not 10 × 9 × 8 × 7 = 5,040.

Time budget and attempt order

P&C contributes around one mark every two papers. That is small, so do not invest more than 75 seconds on a single P&C item in the first pass. If the stem is straightforward — pure nPr, pure nCr, simple word-letter arrangement — take it on the first pass. If it involves overlapping restrictions (always together AND not at the ends) or a long composition stem, flag and move on.

Because P&C is high-confidence when straightforward, getting the one available mark correct often beats spending three minutes on a single complex P&C problem.

Worked AFCAT-style examples

Exam-day strategy

1. Make the order-vs-no-order decision first. If swapping two chosen items gives a new outcome, you need a permutation; otherwise a combination. This single check eliminates the most common P&C error.
2. Use the symmetry identity nCr = nC(n − r) to turn a hard combination into a smaller one. For example, 12C10 is easier as 12C2 = 66.
3. When letters or objects repeat, divide n! by the factorial of each repeat count. Do not skip this step — it is the most common AFCAT P&C trap.
4. For 'always together' use the glue method (tie items into a block, multiply by internal arrangements). For 'never together' use the gap method (arrange the others first, then drop the restricted items into the gaps).
5. For 'at least one' selection problems, use the complement: total minus (none of the required type). It is faster and harder to mis-count than case-by-case addition.
6. Memorise factorials up to 8! and the small nCr values (5C2 = 10, 6C2 = 15, 7C2 = 21, 8C3 = 56, 10C3 = 120). Most AFCAT P&C answers reduce to one of these.
7. Aim for 60–75 seconds on a direct nPr or nCr item, and flag any P&C item that needs more than 90 seconds — the topic contributes too little to fight for two minutes.

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