Question

A committee of 4 people to be formed from a pool of 15 students ( 7 Seniors and 8 Juniors). In how many ways can that be done knowing that every committee must have at least one Senior student?

a) C(14,3)

b) C(14,4) - 1

c) C(7,1) + 3

d) None of the above

I counted the cases:

1 Senior 3 Juniors

2 Seniors 2 Juniors

3 Seniors 1 Junior

4 Seniors 0 Juniors

C(7,1) x C(8,3) + C(7,2) x C(8,2) + C(7,3) x C(8,1) + C(7,4) x C(8,0) = 1295

Am I doing this wrong? I found an answer explaing it like this:

As given one senior is must in committee so taking one senior prior to doing anything. Now we are 14 students ( 6 seniors and 8 juniors) and there are 3 more need to be selected.

selecting 3 from 14 = C(14,3)

Answer 1

The approach is correct

as you mentioned the possible cases for forming the committee with the constraint of 1 senior is are

1 Senior 3 Juniors 7+56 +21 +28 +35 +28 +35+8 + 35+8
2 Seniors 2 Juniors
3 Seniors 1 Junior
4 Seniors 0 Juniors

so 1 senior can be selected in 7c1 ways and 3 juniors can be selected in 8C3

likewise
2 Seniors 2 Juniors
C(7,2) x C(8,2)

3 Seniors 1 Junior
C(7,3) x C(8,1)

4 Seniors 0 Juniors
C(7,4) x C(8,0)

solving the combination formula , we get the final answer as

nCr = n!/r!(n-r)!

= 1295