Given eleven positive real numbers n1 < n2 < n3 < ... < n11, how many triplets can be formed such that in each triplet the first number is less than the second, and the second is less than the third?

Answer Choices

Correct answer marked below

110

165

Correct

220

275

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Official Explanation

Official

To determine the number of triplets from eleven distinct numbers, we can use the combination formula. The number of ways to choose 3 numbers from 11 is given by [m]C^3_{11} =…

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What this question tests

What it tests

Whether you can count arrangements and selections correctly — permutations vs. combinations.

Common trap

Overcounting: treating a selection as ordered when order does not matter, or forgetting to divide by duplicates.