<s>[INST] Combinatorics problem, right solution? We have $6$ lawyers, $7$ engineers and $4$ doctors. We plan on making a committee of $5$ people, and we want at least one person of each profession on board. So for the first place I choose an engineer, for the second a doctor and for the third a lawyer, leaving only $5$ laywers, $6$ engineers and $3$ doctors left.
For the remaining two places, I could have $2$ more people of a single profession. This is $\binom{5}{2}+\binom{6}{2}+\binom{3}{2}$ possibilities.
I could also have two people of different professions; a doctor and a laywer, $\binom{5}{1}\binom{3}{1}$; a doctor and an engineer, $\binom{6}{1}\binom{3}{1}$; or an engineer and a laywer $\binom{6}{1}\binom{5}{1}$.
This adds up to $\binom{5}{2}+\binom{6}{2}+\binom{3}{2}+\binom{5}{1}\binom{3}{1}+\binom{6}{1}\binom{3}{1}+\binom{6}{1}\binom{5}{1}=91$ possible committees.
I have two questions regarding my approach to the problem. Question $a)$ is the reasoning right, am I not overcounting? $b)$ Even if it is right, is there a simpler way to do this? You can see that the sum I end up with, though relatively simple, is quite long and tedious.
Thanks in advance!
 [/INST] There are 6 types of possible committees:
\begin{array} {|r|r|r|}
\hline
3&1&1 \\
\hline
1&3&1 \\
\hline
1&1&3 \\
\hline
2&2&1 \\
\hline
2&1&2 \\
\hline
1&2&2 \\
\hline
\end{array}
For each type of committee, it is necessary to calculate the different groups of people that compose it:
\begin{array} {|c|c|c|c|}
\hline
3&1&1& {6\choose3}{7\choose1}{4\choose1}= 20\cdot7\cdot4 = 560 \\
\hline
1&3&1& {6\choose1}{7\choose3}{4\choose1}= 6\cdot35\cdot4 = 840 \\
\hline
1&1&3& {6\choose1}{7\choose1}{4\choose3}= 6\cdot7\cdot4 = 168 \\
\hline
2&2&1& {6\choose2}{7\choose2}{4\choose1}= 15\cdot21\cdot4 = 1260 \\
\hline
2&1&2& {6\choose2}{7\choose1}{4\choose2}= 15\cdot7\cdot6 = 630 \\
\hline
1&2&2& {6\choose1}{7\choose2}{4\choose2}= 6\cdot21\cdot6 = 756 \\
\hline
\end{array}
Adding 6 gives a total of different committees:
$$ 560+840+168+1260+630+756 = 4214 $$
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