PYQ Vault

NDA Mathematics · Formula sheet

Permutation & Combination formulas

5 formulas and 6 common traps for NDA Mathematics Permutation & Combination, grouped by subtopic.

Full notes with worked examples

Factorials & Binomial Coefficients

Learn this subtopic in the notes

The fundamental principle of counting

Permutations, combinations, and their link

nPr=n!(n−r)!nCr=n!r! (n−r)!nPr= nCr⋅r!^nP_r=\dfrac{n!}{(n-r)!} \qquad ^nC_r=\dfrac{n!}{r!\,(n-r)!} \qquad ^nP_r=\,^nC_r\cdot r!

Factorials: divisibility and trailing zeros

Trailing zeros of n!

Z(n!)=⌊n5⌋+⌊n25⌋+⌊n125⌋+⋯Z(n!)=\left\lfloor\dfrac{n}{5}\right\rfloor+\left\lfloor\dfrac{n}{25}\right\rfloor+\left\lfloor\dfrac{n}{125}\right\rfloor+\cdots

Binomial coefficient identities

nCr= nCn−rnCr+ nCr−1= n+1Cr∑r=0nnCr=2nr⋅nCr=n⋅n−1Cr−1^nC_r=\,^nC_{n-r} \qquad ^nC_r+\,^nC_{r-1}=\,^{n+1}C_r \qquad \sum_{r=0}^{n}{}^nC_r=2^n \qquad r\cdot{}^nC_r=n\cdot{}^{n-1}C_{r-1}

Common traps

Permutation vs combination — does order matter?

Picking a chairperson and a secretary from nn people is a permutation (nP2^nP_2, the two roles are distinct) — but picking a 2-person committee is a combination (nC2^nC_2). Always ask: 'does swapping the two chosen items give a different outcome?' If yes, use nPr^nP_r; if no, use nCr= nPr/r!^nC_r=\,^nP_r/r!.

0!=10!=1, not 00

By definition 0!=10!=1 (it's the empty product, and it makes nCn=n!n! 0!=1^nC_n=\dfrac{n!}{n!\,0!}=1 work). Treating 0!=00!=0 breaks every nC0^nC_0, nCn^nC_n, and boundary case — e.g. the number of ways to choose 0 objects is nC0=1^nC_0=1 (one way: choose nothing), not 00.

Permutations & Restricted Arrangements

Learn this subtopic in the notes

Common traps

Repeated letters → divide by the repeat-factorials

The arrangements of a word with repeated letters is not n!n!. For each letter repeated pp times, swapping those identical copies gives the same word, so you over-count by p!p!. Divide: MATHEMATICS (M, A, T each twice) has 11!2! 2! 2!\dfrac{11!}{2!\,2!\,2!}, not 11!11!.

Combinations & Selections

Learn this subtopic in the notes

Common traps

Use nCr= nCn−r^nC_r=\,^nC_{n-r} — and count the empty set

Symmetry nCr= nCn−r^nC_r=\,^nC_{n-r} means choosing rr to keep equals choosing n−rn-r to leave out — so compute the easier one (50C48= 50C2^{50}C_{48}=\,^{50}C_2). Separately, the number of subsets of an nn-set is 2n2^n, which includes the empty set; 2n−12^n-1 is the count of non-empty subsets only.

'At least one' = total − none (don't sum cases)

For 'at least one X', count the complement: total−(none of X)\text{total}-(\text{none of X}). Summing the cases 'exactly 1, exactly 2, …' is slower and easy to miscount. E.g. at least one typist when choosing 5 from 6 programmers + 4 typists is 10C5− 6C5=252−6=246^{10}C_5-\,^6C_5=252-6=246, not a sum of four separate terms.

Forming Numbers from Digits

Learn this subtopic in the notes

Sum of all numbers formed

Sum of all numbers formed from n distinct digits

Sum=(n−1)!×(sum of digits)×111…1⏟n ones\text{Sum}=(n-1)!\times(\text{sum of digits})\times\underbrace{111\ldots1}_{n\text{ ones}}

Geometric Counting

Learn this subtopic in the notes

Lines, triangles and polygons from points

Geometric counting from points and lines

lines= nC2triangles= nC3diagonals=n(n−3)2parallelograms= mC2⋅nC2\text{lines}=\,^nC_2 \qquad \text{triangles}=\,^nC_3 \qquad \text{diagonals}=\dfrac{n(n-3)}{2} \qquad \text{parallelograms}=\,^mC_2\cdot{}^nC_2

Common traps

Collinear points form no triangle — subtract kC3^kC_3

nC3^nC_3 counts triangles only if no three points are collinear. If kk of the points lie on one line, those kC3^kC_3 triples are degenerate (no triangle), so the answer is nC3− kC3^nC_3-\,^kC_3. Likewise the diagonal formula n(n−3)2= nC2−n\dfrac{n(n-3)}{2}=\,^nC_2-n subtracts the nn sides from all nC2^nC_2 point-pairs.

More NDA Mathematics formula sheets