<s>[INST] 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?
 [/INST] 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*}

One way to count the number of admissible selections is: We select either one girl and three boys, or two girls and two boys or three girls and one boy giving
  \begin{align*}
\binom{5}{1}\binom{6}{3}+\binom{5}{2}\binom{6}{2}+\binom{5}{3}\binom{6}{1}&=20\cdot 5+ 10\cdot 15+ 10\cdot 6\\
&=\color{blue}{310}
\end{align*}

Another way:

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*}

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