CA Foundation · Quantitative Aptitude · Permutations and Combinations
If nC2 = 66 for a positive integer n, what is the value of (n+1)C3?
From nC2 = 66, n(n-1) = 132, so n = 12. Then (n+1)C3 = 13C3 = (13×12×11)/6 = 286. Choosing 220 would result from computing 12C3 instead.
- A286Correct
- B220
- C364
- D312
Explanation
nC2 = n(n-1)/2 = 66 gives n(n-1) = 132, so n = 12. Then 13C3 = (13×12×11)/6 = 286. The value 220 is 12C3, obtained by forgetting to add 1 to n.
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