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Worked Examples · Example 3

Q.Evaluate 8C3^{8}C_{3}.

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Concept understanding — Combinations (nCr)

A combination counts the number of ways to select rr items from nn distinct items when order does not matter — a committee, a hand of cards, a subset.

Selection = combination (order irrelevant); arrangement = permutation (order matters). If the words "choose", "select", or "committee" appear, reach for nCr^nC_r.

How it works

Start from all nPr^nP_r ordered arrangements, then divide out the r!r! orderings within each chosen group that you no longer wish to distinguish.

nCr=n!r! (n−r)!^{n}C_{r}=\frac{n!}{r!\,(n-r)!}

Useful identities:

  • nCr=nCn−r^{n}C_{r}={}^{n}C_{n-r}
  • nCr+nCr−1=n+1Cr^{n}C_{r}+{}^{n}C_{r-1}={}^{n+1}C_{r} (Pascal's rule)
  • nCrnCr−1=n−r+1r\dfrac{^{n}C_{r}}{^{n}C_{r-1}}=\dfrac{n-r+1}{r}

Common problem types

  1. Direct selection — plug into the formula.
  2. With a condition — split into cases (e.g. "exactly 2 women") and multiply the sub-selections.
  3. Find nn and rr — take ratios of consecutive values to kill the factorials.

Quick example

From 7 men and 4 women, form a committee of 5 with exactly 2 women.

  1. Choose 2 women from 4: 4C2=6^{4}C_{2}=6.
  2. Choose the remaining 3 members from the 7 men: 7C3=35^{7}C_{3}=35.
  3. Both must happen, so multiply: 6×35=2106\times 35=\textbf{210} ways.
Watch out

  • Don't use nCr^nC_r when order matters (seating, ranking, forming numbers) — that is a permutation.
  • "At least" conditions need you to add cases (or subtract from the total), not multiply blindly.
  • nC0=1^{n}C_{0}=1 and nCn=1^{n}C_{n}=1 — choosing none or all is a single way.
Tip

For "find n,rn,r" problems, never expand factorials — use nCrnCr−1=n−r+1r\dfrac{^{n}C_{r}}{^{n}C_{r-1}}=\dfrac{n-r+1}{r} to turn consecutive values into two linear equations. (E.g. nCr−1=28, nCr=56, nCr+1=70^{n}C_{r-1}=28,\ ^{n}C_{r}=56,\ ^{n}C_{r+1}=70 gives n=8, r=3n=8,\ r=3.)

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