Question

Let A be the set of first 16 natural numbers. Then the number of ways of selecting 3 numbers from A such that they are all consecutive or none of them are consecutive is

• A. $364$
• B. $378$
• C. $381$
• D. None of these

Answer 1

Answer: (B)

$378$

Let us do the case where they are not consecutive.
Let the 3 numbers be A, B, C.
We place 1 number between A and B and 1 number between B and C.
Thus, $5$ numbers have been used up.
We are left with $11$ numbers to be distributed in $4$ gaps between A, B, C (including the gaps in the beginning and end).
This can be done in ${}^{11+4-1}{C}_{4-1}{=}^{14}{C}_3=364$ ways.
Let us do the case where they are all consecutive.
This can be done as: $(1,2,3);(2,3,4);...(14,15,16)in16-3+1=14$ ways.
Thus, answer $=364+14=378$
Hence, (B) is correct.