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 the third number?

Three numbers can be selected and arranged out of 10 numbers in ¹⁰P₃ ways 10!/7! = 10 x 9 x 8
Now, this arrangement is restricted to a given condition that first number is always less than the second number and second number is always than the third number. Thus, three numbers can be arranged among themselves in 3! ways.
Hence, required number of arrangement = (10 x 9 x 8)/(3 x 2)
= 120 ways