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If there are 10 positive real numbers n1 < n2, n3 .... < n10 How many triplets of these numbers (n1, n2, n3) (n2, n3, n4), ... 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 ?
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- 45
- 90
- 120
- 180
Correct Option: C
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
