Triplets of Real Numbers

CAT 2002 Slot 1 · QA · Easy · Permutations and Combinations

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If there are 10 positive real numbers n1<n2<n3<⋯<n10n_1 < n_2 < n_3 < \dots < n_{10}, how many triplets of these numbers (ni,nj,nk)(n_i, n_j, n_k) can be generated such that in each triplet the first number is always less than the second number, and the second number is always less than the third number?

  1. A.

    45

  2. B.

    90

  3. C.

    120

  4. D.

    180

Answer

C

Explanation

To select a triplet (ni,nj,nk)(n_i, n_j, n_k) with ni<nj<nkn_i < n_j < n_k, any combination of 3 distinct numbers selected from the 10 numbers will uniquely satisfy the given condition when arranged in ascending order. Number of ways = (103)=10×9×83×2×1=120\binom{10}{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120.

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