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CAT 2002 — QA Question 12

Forming a committeeEasy

n1, n2, n3 ... n10 are 10 numbers such that n1 > 0 and the numbers are given in ascending order. How many triplets can be formed using these numbers such that in each triplet, the first number is less than the second number, and the second number is less than the third number?

Answer & solution

  • A

    109

  • B

    27

  • C

    36

  • None of these

Solution

Let us assume that the 10 numbers are 1 - 10.

When the first number is 1,

2nd number as 2, the last number can be chosen in 8 different ways.

2nd number as 3, the last number can be chosen in 7 different ways, and so on.

When 1 is the first number, the possible number of triplets as per the given condition = 8 + 7 + 6 + … + 1 = 36

When the first number as 2, the possible number of triplets will be 7 + 6 + ... + 1 = 28

∴ The total number of triplets = 36 + 28 + 21 + 15 + 10 + 6 + 3 + 1 = 120

Hence, option (d).

CAT 2002 QA Q12: n 1 , n 2 , n 3 ... n 10 are 10 numbers such that n 1 > 0 and the numbers are given in ascending order. How ma — Solution | TheCATExam