Question

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

• A. 45
• B. 90
• C. 120
• D. 180

Hint:

In problems involving combinations, ensure that the order of selection doesn't affect the result.

Answer 1

The Correct Option is B

We need to form triplets of numbers where the numbers are in increasing order. We select 3 numbers from a set of 10 distinct numbers. The number of ways to select 3 distinct numbers from 10 is given by the combination formula: \[ \binom{10}{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120 \] But only half of these triplets will satisfy the condition that the first number is always less than the second, and the second is less than the third. Thus, the number of valid triplets is: \[ \frac{120}{2} = 90 \] Hence, the Correct Answer is 90.