<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] Let $A=\{(l,e,d)\in\{1,2,3,4,5,6\}\times\{1,2,3,4,5,6,7\}\times\{1,2,3,4\}\mid l+e+d=5\}$
Then to be found is $$\sum_{(l,e,d)\in A}\binom{6}{l}\binom{7}{e}\binom{4}{d}$$
Under the sketched conditions for equation $5=l+e+d$ we have the following possibilities:


*

*$5=3+1+1$

*$5=2+2+1$

*$5=2+1+2$

*$5=1+3+1$

*$5=1+2+2$

*$5=1+1+3$
This provides you a view on set $A$ and shows that the summation has $6$ terms.
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