Question 8

If there are 10 positive real numbers $$n_1 < n_2 < n_3 ... < n_{10}$$ , how many triplets of these numbers $$(n_1, n_2, n_3 ), ( n_2, n_3, n_4 )$$ 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?

Solution

For any selection of three numbers, there is only one way in which they can be arranged in ascending order.

So, the answer is $$^{10}C_3 = 120$$

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