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CA Foundation · Quantitative Aptitude · Permutations and Combinations

If (n+1)C3 = 2 × nC2, what is the value of n?

The value of n is 5. Writing (n+1)C3 as (n+1)n(n-1)/6 and equating it with n(n-1), the common factor cancels and leaves n+1 = 6. Check: 6C3 = 20 and 2×5C2 = 20.

  1. A5
  2. B6Correct
  3. C7
  4. D8

Explanation

(n+1)C3 = (n+1)n(n-1)/6 and 2×nC2 = n(n-1). Equating and cancelling n(n-1) (n>1) gives (n+1)/6 = 1, so n = 5? Check: (n+1)/6 = 1 gives n+1 = 6, n = 5. Verify: 6C3 = 20 but 2×5C2 = 20. So n = 5.

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