Question
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 than the third number :(CAT 2002)
Answer: Option C
:
C
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:
C
Ans: (c)
Step 1: select the numbers
Step 2: arrange the selected numbers according to the given condition.
Three numbers can be selected and arranged out of 10 numbers in 10C3 ways = 120.
Now, only one arrangement can be possible as A>B>C, hence the solution will be 120.
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