Permutation of members in a committee

[ritwik06]

Homework Statement

There are 5 girls and 2 boys. A committee of 3 members is to be formed such that there is at least one boy in the committee. Find the number of ways of doing so.

The Attempt at a Solution

One place is fixed for a boy. So choosing one boy out of 2: 2C1
I am left with 2 places and 6(5 girls+1 boy left) people to choose from, therefore number of ways = 6C2
Total ways= 6C2 *2C1
=30

But my book gives me the answer 25. What is wrong with my solution?

 

[Dick]

You choose either 1 boy or 2 boys. Split the cases up.

 

[praharmitra]

you are counting the same case twice. Consider i choose boy A first(in the 2C1 way), and then while choosing two from the other 6 i choose boy B, and a girl.

In another way, i can choose boy B first, and then A and a girl. they both give the same selection, but you are counting both.

Subtract from your answer all those cases in which both boys are there. does it give you your answer??

 

[ritwik06]

You choose either 1 boy or 2 boys. Split the cases up.

you are counting the same case twice. Consider i choose boy A first(in the 2C1 way), and then while choosing two from the other 6 i choose boy B, and a girl.

In another way, i can choose boy B first, and then A and a girl. they both give the same selection, but you are counting both.

Subtract from your answer all those cases in which both boys are there. does it give you your answer??

Thank you very much. I now understand my folly.