Vicky wrote:

oh man.. silly mistake...

thanks for correction.

Second solution:

4C1 = number of ways one senior can be selcted

9C2 = number of ways any 2 people can be selected from remaining 9 people.

Thus ways to make groups of 3 people groups have atleast 1 senior.

= 4C1*9C2

hope this clarifies.

1. C ...both are required to solve the problem

2. 100 is the answer..

Most simplest way is 10C3 - 6C3

Vicks...did you notice the ATLEAST one senior part in the problem

Question, why do you multiply the combinations?