Problem on Basic Combinations

Question

A debate club consists of $6$ girls and $4$ boys. A team of $4$ members is to be selected from this club including the selection of a captain (from among these $4$ members) for the team. If the team has to include at most one boy, then the number of ways of selecting the team is?

My method was to use combinations: $$\frac{4!}{3!} \cdot \frac{6!}{3!\,3!} + \frac {6!}{4!\,2!} = 95$$ But that is not the right answer. Where have I gone wrong?

In your method, you are not counting the number of choices for choosing a captain. — Anurag A, Mar 13 '17 at 18:27
You have ignored that one of the selected members will be the captain. — Evangelos Bampas, Mar 13 '17 at 18:28
I get it. There were 4 ways to select a captain amongst them, so the answer in effect would be 4(95) = 380. — Karthik, Mar 13 '17 at 18:32
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score 1 · Accepted Answer · answered Mar 13 '17 at 18:31

You have added the ways of choosing one boy and the ways of choosing 4 girls ignoring the boys; and in neither case have you distinguished different choices for the captain.

Your breakdown between 1 boy and 2 boys is a good start.

One boy gives you $\binom41$ choices for the boy, $\binom63$ choices for the girls, and $4$ choices for who is the captain.

No boys gives you $\binom64$ choices for the girls, times $4$ choices for who is the captain.

The total is $320+60 = 380$ team selections.

Problem on Basic Combinations

1 Answers1