CA Foundation 2026 · Paper 3 · Quantitative AptitudeQ69 · 1 mark↻ Appears in 5 of 6 yearsDefective question — grace marks awarded
Four cards are drawn at random from a standard deck of 52 playing cards without replacement. In how many ways it can be done such that the selected cards consist of exactly one Jack and three Aces?
Figure for question 69
  1. (A) 2304
  2. (B) 2440
  3. (C) 2260
  4. (D) 2164

The exam board itself declared this question defective — see the Answer tab. None of the four options above is the correct answer.

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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. …

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