Permutations and Combinations

What This Topic Covers

The NDA syllabus groups this chapter under Algebra and it tests three overlapping ideas: counting principles, permutations, and combinations. You need to be comfortable switching between them in a single question.

The topic also connects directly to Probability — sample spaces are often computed using \(^{n}C_{r}\) — and to Binomial Theorem, where the binomial coefficients are combination values. Understanding this chapter well gives you a compounding advantage across multiple topics.

Exam Pattern & Weightage

The table below is built entirely from the PYQ file. Each row shows the approximate question count for that year's paper.

Core Concepts

Factorial

The factorial \(n!\) is the product of all positive integers from 1 to \(n\). By definition, \(0! = 1\). Factorials grow extremely fast: \(10! = 3{,}628{,}800\). The number of trailing zeros in \(n!\) equals the number of times 5 is a factor, calculated as \(\lfloor n/5 \rfloor + \lfloor n/25 \rfloor + \lfloor n/125 \rfloor + \cdots\)

Permutations — \(^{n}P_{r}\)

A permutation is an ordered arrangement. When you pick \(r\) objects from \(n\) distinct objects and the order in which you pick them matters, you use \(^{n}P_{r}\). For instance, seating 6 girls in 2 chairs is \(^{6}P_{2} = 6 \times 5 = 30\) — exactly the answer from the 2011-I paper.

Special case: arranging all \(n\) distinct objects gives \(^{n}P_{n} = n!\). So 6 distinct books on a shelf can be arranged in \(6! = 720\) ways.

Combinations — \(^{n}C_{r}\)

A combination is an unordered selection. When the order does not matter, you use \(^{n}C_{r}\). Selecting a cricket team of 11 from 15 players is \(^{15}C_{11} = 1365\) — from the 2014-II paper.

This relationship was directly tested in 2013-I: if \(^{77}P_{31} = x\) and \(^{77}C_{31} = y\), then \(x = 31! \cdot y\), so \(x > y\). The answer is option (d).

Important Combination Properties

This rule cracks open the entire 2020-I question — \(^{20}C_{n+2} = {}^{20}C_{n-2}\) implies \((n+2) + (n-2) = 20\), so \(n = 10\). No factorial expansion needed.

The 2011-I paper asked for the total number of combinations of \(n\) different things taken \(1, 2, 3, \ldots, n\) at a time. The answer is \(2^{n} - 1\) (subtracting \(^{n}C_{0} = 1\) for the empty selection).

This appeared in 2022-I. The sum equals \((1+2)^{n} = 3^{n}\) by the binomial theorem with \(x = 2\).

Permutations with Repeated Letters

When objects repeat, divide by the factorial of each repetition count. BASEBALL has 8 letters — 2 B's, 2 A's, 2 L's — giving \(\frac{8!}{2! \cdot 2! \cdot 2!} = 5040\) arrangements (2012-II). PERMUTATION has 11 letters with T repeated twice: \(\frac{11!}{2!} = 19{,}958{,}400\) arrangements (2017-II).

Circular Permutations

When objects are arranged in a circle, one position is fixed to remove rotational equivalence. For \(n\) distinct objects in a circle: \((n-1)!\) arrangements. The tea party seating question (2017-II) uses this principle extended to two sides of a table.

The necklace divisor of 2 accounts for clockwise vs anti-clockwise being the same arrangement when the necklace is flipped. A round table with no flip uses \((n-1)!\); a beaded necklace uses \(\frac{(n-1)!}{2}\).

Distribution Problems

When distributing \(n\) distinct objects into \(r\) distinct boxes (or sending \(n\) letters into \(r\) letterboxes) with no restriction, each object independently goes to any box: \(r^{n}\) ways. Posting 5 letters into 7 letterboxes = \(7^{5} = 16{,}807\) (2014-II). Giving 3 holiday tickets to 20 employees where each can get any number = \(20^{3} = 8{,}000\) (2015-I).

Geometric Applications

These appear almost every year and follow fixed patterns:

The 2015-II paper gave 44 diagonals and asked for the number of sides. Setting \(\frac{n(n-3)}{2} = 44\) gives \(n^{2} - 3n - 88 = 0\), so \((n-11)(n+8) = 0\), meaning \(n = 11\). Similarly in 2025-I, 20 diagonals gives \(n^{2} - 3n - 40 = 0\), so \(n = 8\).

Constrained Selections

These are some of the most frequent question types. Always include a specific item: reduce \(n\) by 1 and choose \(r-1\) from the rest. Always exclude: reduce \(n\) by 1 and choose \(r\) from the rest.

Selecting 5 from 8 players excluding one particular player (2021-I): that player is out, so choose 5 from the remaining 7 = \(^{7}C_{5} = 21\).

If captain is always included in a team of 11 from 15 (2015-II): captain is fixed, choose 10 from remaining 14 = \(^{14}C_{10} = 1001\).

Worked Examples

Example 1 — Finding n from a P-ratio (2011-I)

Question: What is the value of \(n\) if \(^{15}P_{n-1} : {}^{16}P_{n-2} = 3 : 4\)?

• Write \(^{15}P_{n-1} = \dfrac{15!}{(16-n)!}\) and \(^{16}P_{n-2} = \dfrac{16!}{(18-n)!}\).
• Set up the ratio: $$\frac{15!/(16-n)!}{16!/(18-n)!} = \frac{3}{4}.$$
• Simplify: $$\frac{(18-n)!}{16 \cdot (16-n)!} = \frac{3}{4}.$$
• Since \((18-n)! = (18-n)(17-n)(16-n)!\), this gives \((18-n)(17-n) = 12\).
• Expand: \(n^{2} - 35n + 294 = 0\), so \((n-14)(n-21) = 0\).
• Since \(n \le 15\) (so that \(^{15}P_{n-1}\) is defined), \(n = 14\). Answer: (c).

Example 2 — Words with repeated letters: GLOOMY (2013-I)

Question: In how many ways can the letters of GLOOMY be arranged so that the two O's are not together?

• Total letters in GLOOMY = 6, with O repeated twice. Total arrangements = \(\dfrac{6!}{2!} = 360\).
• Treat the two O's as a single unit. Now 5 units (OO, G, L, M, Y) arranged in \(5! = 120\) ways.
• Inside the unit, the two O's are identical, so no extra factor needed.
• Arrangements with O's together = 120.
• Arrangements with O's NOT together = \(360 - 120 = 240\). Answer: (a).

Example 3 — Geometric: triangles from PQR sides (2024-I)

Question: Triangle PQR has 3 points on PQ, 4 on QR, and 5 on RP. How many triangles can be formed using these points as vertices?

• Total points = \(3 + 4 + 5 = 12\).
• Total ways to choose 3 points = \(^{12}C_{3} = 220\).
• Points on the same side are collinear so they cannot form a triangle. Subtract degenerate selections.
• Collinear sets: \(^{3}C_{3} + {}^{4}C_{3} + {}^{5}C_{3} = 1 + 4 + 10 = 15\).
• Valid triangles = \(220 - 15 = 205\). Answer: (a).

Example 4 — Rank of word in dictionary (2015-II)

Question: Letters of AGAIN arranged alphabetically — what word is at the 50th position?

• Letters in sorted order: A, A, G, I, N. Note A repeats twice.
• Words starting AA: \(\dfrac{3!}{1!} = 6\). Words starting AG: \(\dfrac{3!}{2!}\) (two A's) … actually \(3! = 6\). Total AA = 6, AG = 6, AI = 6, AN = 6, GA = 6, GI = 3 (two equal letters AA), GN = 3, IA = 6, IG = 3, IN = 3, NA = 6. Cumulative up through NA = 54.
• After the first 48 words (through IN = 3 gives cumulative 48), words starting NA begin.
• 49th word = NAAGI, 50th word = NAAIG. Answer: (b).

Example 5 — \(^{n}C_{4}, {}^{n}C_{5}, {}^{n}C_{6}\) in AP (2021-II)

Question: If \(^{n}C_{4}, {}^{n}C_{5}\) and \(^{n}C_{6}\) are in AP, what is \(n\)?

• AP condition: \(2 \cdot {}^{n}C_{5} = {}^{n}C_{4} + {}^{n}C_{6}\).
• Divide through by \(^{n}C_{4}\): $$2 \cdot \frac{{}^{n}C_{5}}{{}^{n}C_{4}} = 1 + \frac{{}^{n}C_{6}}{{}^{n}C_{4}}.$$
• \(\dfrac{{}^{n}C_{5}}{{}^{n}C_{4}} = \dfrac{n-4}{5}\) and \(\dfrac{{}^{n}C_{6}}{{}^{n}C_{4}} = \dfrac{(n-4)(n-5)}{30}\).
• So \(\dfrac{2(n-4)}{5} = 1 + \dfrac{(n-4)(n-5)}{30}\). Multiply by 30: \(12(n-4) = 30 + (n-4)(n-5)\).
• Let \(m = n-4\): \(12m = 30 + m(m-1)\). Gives \(m^{2} - 13m + 30 = 0\), so \((m-3)(m-10) = 0\).
• Factoring the original cubic gives \(n = 7\) as the valid answer. Answer: (a).

Example 6 — Identical-letter arrangement: ASSASSINATION

Question: How many distinct words can be formed using all the letters of the word ASSASSINATION?

• Count the letters: A, S, S, A, S, S, I, N, A, T, I, O, N — total \(n = 13\).
• Identify repetitions: A appears 3 times, S appears 4 times, I appears 2 times, N appears 2 times. T and O appear once each.
• Apply the repeated-letters formula: $$\text{Arrangements} = \frac{13!}{3! \cdot 4! \cdot 2! \cdot 2!}.$$
• Numerator \(13! = 6{,}227{,}020{,}800\). Denominator \(= 6 \cdot 24 \cdot 2 \cdot 2 = 1{,}728\).
• Final count \(= 6{,}227{,}020{,}800 / 1{,}728 = 3{,}603{,}600\) distinct arrangements.

Example 7 — Equality Rule: solving for r

Question: If \(^{15}C_{3r} = {}^{15}C_{r+3}\), find the value of \(r\).

• By the Equality Rule, either \(3r = r + 3\) OR \(3r + (r+3) = 15\).
• Case 1: \(2r = 3 \implies r = 1.5\) — rejected, \(r\) must be a non-negative integer.
• Case 2: \(4r + 3 = 15 \implies 4r = 12 \implies r = 3\).
• Verify: \(^{15}C_{9} = {}^{15}C_{6}\) — true by symmetry. So \(r = 3\).

Example 8 — "Always together" arrangement

Question: In how many ways can the letters of EQUATION be arranged so that all the vowels are together?

• EQUATION has 8 distinct letters. Vowels: E, U, A, I, O (5 vowels). Consonants: Q, T, N (3 consonants).
• Treat all 5 vowels as a single block. Now arrange the block plus 3 consonants → 4 units in \(4! = 24\) ways.
• Inside the block, the 5 vowels can be permuted among themselves in \(5! = 120\) ways.
• Total arrangements with vowels together = \(4! \times 5! = 24 \times 120 = 2{,}880\).
• The "block" or "glue" trick is the standard NDA approach for any "X items always together" question.

Exam Shortcuts (Pro-Tips)

Permutations and Combinations questions are usually time-sinks if you set them up from first principles. The shortcuts below collapse standard sub-types into one-line answers — every one of them maps to an NDA pattern seen across 2010–2025.

Shortcut 1 — Circular arrangement is \((n-1)!\)

For \(n\) distinct objects around a round table, fix one seat to eliminate rotational duplicates. That leaves \((n-1)!\) arrangements. For a necklace or garland, divide further by 2 because clockwise and anti-clockwise look identical when flipped.

Shortcut 2 — Identical objects: divide by each repetition's factorial

Any time you arrange \(n\) items where some are alike, divide \(n!\) by the factorial of each repetition count. This single rule covers every BASEBALL / MATHEMATICS / ASSASSINATION-style question.

Shortcut 3 — Pascal identity collapses adjacent C-values

Whenever a question mixes \(^{n}C_{r}\) and \(^{n}C_{r-1}\) in a sum, fold them into a single \(^{n+1}C_{r}\) value. This is the fastest route to handle sums or AP-condition questions involving consecutive combinations.

Shortcut 4 — Restricted positions: "always included" and "always excluded"

If \(p\) particular objects must always be included in a selection of \(r\) from \(n\), those \(p\) slots are pre-filled — just choose the remaining \((r-p)\) from \((n-p)\). If \(p\) objects must always be excluded, treat them as removed before selecting.

Example: A team of 11 from 15 with captain always included → \(^{14}C_{10} = 1001\). A team of 5 from 8 with one player excluded → \(^{7}C_{5} = 21\). Both have appeared verbatim in NDA papers (2015-II, 2021-I).

Shortcut 5 — "At least one" means use the complement

Selection questions asking for "at least one of type X" balloon if you sum cases for exactly 1, 2, 3 … Use the complement instead — subtract the count of "none chosen" from the total.

Example: from 6 men and 4 women, the number of committees of 3 with at least one woman = \(^{10}C_{3} - {}^{6}C_{3} = 120 - 20 = 100\). Faster than enumerating 1W+2M, 2W+1M, 3W separately.

Shortcut 6 — Diagonals and triangles direct formulas

For any \(n\)-sided polygon, memorise diagonals \(= \dfrac{n(n-3)}{2}\). For "triangles from \(n\) points, \(m\) collinear" memorise \(^{n}C_{3} - {}^{m}C_{3}\). For "straight lines from \(n\) points, \(m\) collinear" memorise \(^{n}C_{2} - {}^{m}C_{2} + 1\).

Shortcut 7 — Derangements (the "wrong envelope" problem)

If \(n\) letters go into \(n\) addressed envelopes such that no letter sits in its correct envelope, the count is the derangement \(D_{n}\). Memorise the small values; NDA has never tested \(n > 5\) for this sub-type.

Closed-form: $$D_{n} = n!\left(1 - \tfrac{1}{1!} + \tfrac{1}{2!} - \tfrac{1}{3!} + \cdots + \tfrac{(-1)^{n}}{n!}\right).$$ For NDA, the three values above are enough.

Shortcut 8 — Group division formula

To split \((m + n)\) distinct objects into two unequal groups of size \(m\) and \(n\): \(\binom{m+n}{m} = \dfrac{(m+n)!}{m!\,n!}\). If the two groups are of equal size \(n\), divide by an extra \(2!\) because the groups themselves are interchangeable.

Common Question Patterns

After reviewing every question from 2010 through 2025, six recurring patterns emerge. Learning to spot the pattern in the first reading saves 30–60 seconds per question.

Preparation Strategy

Permutations and Combinations rewards structured preparation more than most topics. Here is a week-by-week plan based on NDA PYQ analysis.

Week 1 — Build the formula base

Master the definitions of \(^{n}P_{r}\) and \(^{n}C_{r}\). Practice computing values by hand: \(^{6}P_{2}\), \(^{15}C_{11}\), \(^{7}C_{3} \times {}^{5}C_{2}\). Make sure you can quickly spot whether a question needs permutation (order matters) or combination (order does not matter). A good test: if you are selecting a team, order doesn't matter — use C. If you are arranging people in chairs or letters in words, order matters — use P.

Week 2 — Word arrangement questions

This sub-type appears in almost every NDA paper. Practice the following progression:

• Count arrangements of a word with no repeats (TIGER = \(5! = 120\)).
• Count arrangements with repeated letters (BASEBALL = \(\dfrac{8!}{2! \cdot 2! \cdot 2!} = 5040\)).
• Apply constraints: vowels together (treat as block), vowels not together (complement), vowels in specific positions.
• Rank in dictionary order (AGAIN 50th word = NAAIG).

These questions connect naturally to Sets, Relations and Functions — the relation count question in 2015-I (distinct relations from B to A = \(2^{n(A) \times n(B)} = 2^{12} = 4096\)) uses a similar counting approach.

Week 3 — Geometric counting

Drill the three formulas: lines, triangles, diagonals. Then practice questions with collinear points (subtract degenerate cases). Move on to the parallelogram formula \(^{m}C_{2} \times {}^{n}C_{2}\). NDA has tested this every 2–3 years; a question from 2023-I gave 60 parallelograms from \(m\) and \(n\) parallel lines, leading to \(m = 5, n = 4, m + n = 9\).

Week 4 — Mixed PYQ practice and speed

At this stage, stop doing concept practice and shift to timed PYQ solving. Target 2 minutes per question. Common slow spots: setting up the constraint (always include/exclude), handling the collinear point subtraction, and expanding word arrangement constraints. Time yourself on 10-question sets from this topic alone, then review every mistake.

Cross-topic connections worth reinforcing: this topic feeds directly into Probability (favorable outcomes / total outcomes both use \(^{n}C_{r}\)), into Sequences and Series (the \(a_{n} = n(n!)\) question from 2021-II telescopes into \(11! - 1\)), and into Binomial Theorem (binomial coefficients = \(^{n}C_{r}\) values). Solving the cross-topic links in one sitting consolidates memory better than isolated drill.

For chapter-level context, return to NDA Maths to see how Permutations and Combinations fits within the full syllabus and mark scheme.

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