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Non-Adjacent Pairs Singing in Circle

CAT 2004 Slot 1 · Quantitative Ability · Medium · Permutations and Combinations

This is a medium Quantitative Ability question from the CAT 2004 Slot 1 paper. It tests Permutations and Combinations. The full answer key and a step-by-step explanation are below — try it yourself first, then reveal the solution.

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NN persons stand on the circumference of a circle at distinct points. Each possible pair of persons, not standing next to each other, sings a two-minute song one pair after the other. If the total time taken for singing is 2828 minutes, what is NN?

  1. A.

    55

  2. B.

    77

  3. C.

    99

  4. D.

    None of the above

Answer

B

Explanation

Total pairs of NN persons is (N2)=N(N−1)2\binom{N}{2} = \frac{N(N-1)}{2}. The number of adjacent pairs is NN. Thus, number of non-adjacent pairs =N(N−1)2−N=N(N−3)2= \frac{N(N-1)}{2} - N = \frac{N(N-3)}{2}. Each song takes 2 minutes, so total time is: 2×N(N−3)2=N(N−3)=282 \times \frac{N(N-3)}{2} = N(N-3) = 28 N2−3N−28=0  ⟹  (N−7)(N+4)=0N^2 - 3N - 28 = 0 \implies (N - 7)(N + 4) = 0 Since N>0N > 0, N=7N = 7.

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