# Committee selection:Where is my mistake?

There are $5$ boys and $6$ girls. A committee of $4$ is to be selected so that it must consist at least one boy and at least one girl?

I consider $1$ boy can be chosen in $5$ ways.

Also, consider $1$ girl can be chosen in $6$ ways.

So, others $2$ can be chosen in $9C2$ ways.

So, required answer is $5 * 6 * 9C2$

But this is the wrong answer. Where is my mistake?

• Please read the (self-learning) tag description more closely "The process of studying mathematics without formal instruction. Don't use this tag just because you were self-studying when you came across the mathematical question you're asking; it is only for when the fact that you're self-studying is what your question is about. " – Trevor Gunn Jul 20 '17 at 13:28
• You've overcounted. Let's number the boys B1-B5, and the girls G1-G6. Suppose you from the committee your way: we pick B3, then G2, then B5, then G3. But, you're counting separatedly the committee formed of B5, then G3, then B3, then G2. – Mees de Vries Jul 20 '17 at 13:28
• Probably, for this question, it is easiest to consider the cases "1 boy, 3 girls" and "2 boys, 2 girls" and "3 boys, 1 girl" separately. – Mees de Vries Jul 20 '17 at 13:29
• Alternatively you can count how many committees have no boys or have no girls in it. – drhab Jul 20 '17 at 14:06

Let's denote the girls $\{g_1,g_2,\ldots,g_6\}$ and the boys $\{b_1,b_2,\ldots,b_5\}$.
You count some configurations more than once. The selection $g_1,g_2,b_1,b_2$ is counted four times namely as \begin{align*} (g_1,b_1),(g_2,b_2)\\ (g_1,b_2),(g_2,b_1)\\ (g_2,b_1),(g_1,b_2)\\ (g_2,b_2),(g_1,b_1) \end{align*}
We count all selections of $4$ people and subtract the selections containing only girls and only boys giving \begin{align*} \binom{11}{4}-\binom{5}{4}-\binom{6}{4}&=330-5-15\\ &=\color{blue}{310} \end{align*}