CA Foundation · Quantitative Aptitude · Permutations and Combinations
If (n+1)P3 = 6 × nP2 holds for a positive integer n, what is the value of n?
The answer is 5. Writing (n+1)P3 as (n+1)n(n−1) and nP2 as n(n−1), the common factor cancels and leaves n+1 = 6, so n = 5. Verification: 6P3 = 120 equals 6 × 5P2 = 6 × 20 = 120.
- A4
- B5Correct
- C6
- D7
Explanation
(n+1)P3 = (n+1)n(n−1) and nP2 = n(n−1). The equation gives (n+1)n(n−1) = 6n(n−1), so n+1 = 6 and n = 5. Check: 6P3 = 120 and 6 × 5P2 = 6 × 20 = 120. Option 6 comes from setting n = 6 directly instead of n+1 = 6.
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