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 $$^{10}C_3 = 120$$

Video Solution

video

Create a FREE account and get:

  • All Quant CAT complete Formulas and shortcuts PDF
  • 35+ CAT previous year papers with video solutions PDF
  • 5000+ Topic-wise Previous year CAT Solved Questions for Free

cracku

Boost your Prep!

Download App