Permutations and Combinations for NDA Mathematics

Concept

Permutations and combinations sit at the heart of combinatorics — the branch of mathematics that counts arrangements and selections. If you've ever wondered how many ways a commanding officer can assign 3 posts from 10 soldiers, or how many 4-digit codes avoid repetition, this is the chapter that answers it.

Here is the core distinction, and it is the one students blur most often:

• Permutation — order matters. The arrangement AB is different from BA.
• Combination — order does not matter. The selection {A, B} is the same as {B, A}.

A useful analogy: think of forming a military squad versus forming a firing line. Picking 5 soldiers for a patrol squad is a combination — it doesn't matter who is listed first on the roll. But arranging those same 5 soldiers into a single-file march past is a permutation — position number one is distinct from position number five.

The mathematical machinery behind both is the factorial. For any positive integer $n$:

$n! = n \times (n-1) \times (n-2) \times \cdots \times 2 \times 1$

By convention, $0! = 1$ — this is not arbitrary; it makes the formulas consistent.

From the factorial, two formulas follow directly:

$^nP_r = \frac{n!}{(n-r)!} \quad \text{(arrangements of } r \text{ from } n\text{)}$

$^nC_r = \frac{n!}{r!(n-r)!} \quad \text{(selections of } r \text{ from } n\text{)}$

Notice that $^nP_r = r! \times {^nC_r}$. The factor of $r!$ is exactly the number of ways to arrange a selected group — so permutation is just combination multiplied by the internal arrangements. Keep this relationship in your head; it resolves half the confusion instantly.

NDA papers test this chapter consistently. Questions are usually not raw formula application — they involve constraints: "leading digit cannot be zero", "vowels must stay together", "consonants keep their positions". The formula is never the hard part. Reading the constraint carefully is.

Deep Dive

Fundamental Counting Principle

Before any formula, you need the multiplication and addition principles:

• Multiplication principle: If task A can be done in $m$ ways and task B (independent) in $n$ ways, both together can be done in $m \times n$ ways.
• Addition principle: If task A can be done in $m$ ways OR task B in $n$ ways (mutually exclusive), the total is $m + n$ ways.

Most NDA problems are multiplication principle problems with one tricky constraint thrown in.

Permutations

All $n$ distinct objects: $n!$ arrangements.

$r$ objects from $n$ distinct objects: $^nP_r = \frac{n!}{(n-r)!} = n(n-1)(n-2)\cdots(n-r+1)$

The second form (product of $r$ consecutive integers descending from $n$) is faster to evaluate mentally than the fraction.

Permutations with repetition in the collection: If you have $n$ objects where one type repeats $p$ times, another repeats $q$ times, etc., the number of distinct arrangements is: $\frac{n!}{p!\, q!\, \cdots}$

This formula appears whenever a word has repeated letters — a very common NDA setup.

Circular permutations: Arranging $n$ distinct objects in a circle gives $(n-1)!$ arrangements, because one object can be fixed to eliminate rotational equivalence. If the arrangement can also be flipped (like a necklace or bracelet), divide further by 2: $\frac{(n-1)!}{2}$.

Combinations

$^nC_r = \frac{n!}{r!(n-r)!}$

Key properties you must know cold:

1. $^nC_r = {^nC_{n-r}}$ — choosing $r$ to include is the same count as choosing $n-r$ to exclude.
2. $^nC_0 = {^nC_n} = 1$.
3. $^nC_r + {^nC_{r-1}} = {^{n+1}C_r}$ — Pascal's identity, useful when $n$ or $r$ appear as unknowns.
4. If $^nC_r = {^nC_s}$, then either $r = s$ or $r + s = n$.

Property 4 is a direct NDA question type — see the PYQ on $^{5n}C_r = {^{5n}C_{n+r}}$.

Diagonals of a Polygon

A polygon with $n$ sides has $n$ vertices. Any 2 vertices define a line segment. Total line segments = $^nC_2$. Subtract $n$ sides to get diagonals:

$\text{Diagonals} = \frac{n(n-1)}{2} - n = \frac{n(n-3)}{2}$

This is a formula NDA recycles frequently. Memorise it. To reverse-engineer $n$ from a given diagonal count, set up a quadratic and solve.

Stars and Bars (Integer Solutions)

The number of positive integer solutions to $x_1 + x_2 + \cdots + x_k = n$ is $^{n-1}C_{k-1}$.

The number of non-negative integer solutions is $^{n+k-1}C_{k-1}$.

The logic: positive-integer case. Give each variable 1 unit first (to ensure positivity), then distribute the remaining $n - k$ units freely. But the neater way to remember it — imagine placing $n$ stars in a row and inserting $k-1$ dividers among the $n-1$ gaps. That is $^{n-1}C_{k-1}$ ways.

Constrained Digit Problems

When forming numbers with digit constraints:

1. Fill the most constrained position first — usually the leading digit (cannot be zero).
2. Then fill remaining positions in decreasing constraint order.
3. Use multiplication principle across positions.

This is the approach that prevents the classic error of forgetting that zero cannot lead a multi-digit number.

"At Most / At Least" Selections

• At most $r$ things from $n$: Sum $^nC_0 + {^nC_1} + \cdots + {^nC_r}$. Watch whether the problem intends to include the empty selection ($^nC_0 = 1$) or not.
• At least $r$: Sum from $r$ to $n$, or equivalently total $2^n$ minus the "at most $r-1$" count.

Memory Tricks & Shortcuts

Fast-Solving Framework

When you see a P&C question in the exam hall, run this decision tree before writing anything:

Step 1 — Order or no order? Words like arrange/sequence/rank → permutation. Words like select/choose/committee → combination.

Step 2 — Any constraint? Leading digit zero forbidden → fill leading digit first. Letters must stay together → use block method. Positions of a type are fixed → permute each type independently and multiply.

Step 3 — Repetition in the pool? Repeated letters/digits in the source → divide by factorial of each repeated count.

Step 4 — "At most" / "At least" / "Exactly"? Exactly $r$ → single $^nC_r$ term. At most $r$ → sum from 0 to $r$ (check whether empty set counts). At least $r$ → sum from $r$ to $n$ or use complement.

Step 5 — Does it match a known formula? Polygon diagonals, integer-solution counting (stars and bars), circular permutations — if the problem matches a known template, apply the formula directly without re-deriving.

Total decision time: under 20 seconds. Then execute the arithmetic cleanly.

Solved PYQs

Why this question: The most common NDA digit-question trap — forgetting that zero cannot occupy the leading position. This appears nearly every year in some form.

Solving path: Digits available: {0, 1, 2, 3}. Target: 4-digit numbers greater than 1000. Since we have exactly 4 digits and no repetition, every valid arrangement uses all 4 digits — so we need all 4-digit arrangements of {0,1,2,3} that don't start with 0. Thousands place: 3 choices (1, 2, 3). Remaining 3 places: fill with the 3 leftover digits in any order → $3! = 6$ ways. Total = $3 \times 6 = 18$.

Why this question: Tests the diagonal formula — a formula NDA recycles, and the reverse problem (given diagonals, find $n$) is the harder direction that eliminates unprepared students.

Solving path: Use the diagonal formula $D = \frac{n(n-3)}{2}$. Set $\frac{n(n-3)}{2} = 20$, so $n(n-3) = 40$, giving $n^2 - 3n - 40 = 0$. Discriminant: $9 + 160 = 169 = 13^2$. So $n = \frac{3 + 13}{2} = 8$. Verify: $\frac{8 \times 5}{2} = 20$. Confirmed.

Why this question: Teaches the "positions are fixed, types permute independently" principle — a cleaner and faster approach than brute-force listing.

Solving path: DELHI: D(1), E(2), L(3), H(4), I(5). Vowels occupy positions 2 and 5: {E, I} can be arranged in those 2 positions in $2! = 2$ ways. Consonants occupy positions 1, 3, 4: {D, L, H} can be arranged in those 3 positions in $3! = 6$ ways. Total = $2 \times 6 = 12$.

Why this question: Stars-and-bars is a direct formula question. NDA occasionally drops one in to see if you know the template, or whether you'll waste time listing solutions.

Solving path: Positive integer solutions to $x + y + z = n$ count as $^{n-1}C_{2}$. Here $n = 5$: $^{4}C_{2} = \frac{4 \times 3}{2} = 6$. Quick verification by listing: $(1,1,3),(1,3,1),(3,1,1),(1,2,2),(2,1,2),(2,2,1)$ — exactly 6.

Why this question: Tests the complement property of combinations. Students who don't know $^nC_r = {^nC_{n-r}}$ will expand factorials and grind for 3 minutes. Students who do will finish in 15 seconds.

Solving path: $^{5n}C_r = {^{5n}C_{n+r}}$. If $r \neq n+r$ (which they aren't, since $n > 0$), the complement property gives $r + (n+r) = 5n$. So $2r + n = 5n$, meaning $2r = 4n$, thus $r = 2n$.

Common Mistakes

• Forgetting the empty selection. "At most 3 from 6" — many students compute $^6C_1 + {^6C_2} + {^6C_3} = 41$ and stop. But "at most 3" includes selecting 0 things ($^6C_0 = 1$), giving 42. Read whether the problem context allows an empty selection, and include it when it does.

• Zero in the leading position. In digit problems with zero as one of the options, every student knows zero can't lead — and then half of them still compute $n!$ and forget. The fix is to always fill the leading position first, not last.

• Permutation when combination is needed (and vice versa). The word "ways" alone tells you nothing. "In how many ways can a committee of 3 be formed" is combination. "In how many ways can a president, vice-president, and secretary be chosen" — same 3 people, but now it's permutation. The role differentiation is the signal.

• Repeated letters ignored. If a word has repeated letters and you arrange it, you must divide by the factorial of each repeated letter's count. CAPITAL has two A's — so any arrangement count gets divided by $2!$. Missing this inflates your answer by the repetition factor.

• Circular vs linear permutation confusion. A circular arrangement of $n$ people is $(n-1)!$, not $n!$. Students who linearise a circular problem overcount by a factor of $n$.

• Stars and bars: positive vs non-negative solutions. $^{n-1}C_{k-1}$ is for positive integer solutions. For non-negative, it's $^{n+k-1}C_{k-1}$. The NDA question will specify "positive integers" — don't apply the non-negative formula by default.

Related Topics

• probability
• binomial theorem
• mathematical induction
• sets and relations
• sequences and series
• matrices and determinants

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