<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] For groups of $6, 7$, and $4$ distinct types of members, would the counting be easier by counting all possible committees while ignoring the constraint, then subtracting the non-valid committees?
$$\binom{6 + 7 + 4}{5} - \left[ \binom{6+7}{5}+\binom{7+4}{5}+ \binom{6+4}{5} \right]$$
Non-valid committees are those which entirely omit one of the types of members.
(Ah: This doesn't work: The subtracted count is too high, as it duplicates non-valid committees.  Trying again ...
$$\binom{6+7+4}{5}-
\left[
\left[ \binom{6+7}{5} - \left[\binom{6}{5}+\binom{7}{5}\right] \right] +
\left[ \binom{7+4}{5} - \left[\binom{7}{5}+\binom{4}{5}\right] \right] +
\left[ \binom{6+4}{5} - \left[\binom{6}{5}+\binom{4}{5}\right] \right]
\right] -
\left[ \binom{6}{5} + \binom{7}{5} + \binom{4}{5} \right]$$
The number of committees of five members, minus the number of committees with exactly two types of members, minus the number of committees with exactly one type of member.
For two given types of members, committees with exactly two types of members are all committees of those two types of members, minus committees of only one of the types of members, minus committees of only the other type of members.
To show the complete expression, I've left the terms which evaluate to 0, and haven't simplified.
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