Ttyjukl

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!

I think Inclusion Exclusion is an easier approach.

If we ignore the restriction, there are $binom {17}5$ ways to choose the group.

We then exclude the choices which miss one specified profession. That's an exclusion of $$binom {11}5+binom {10}5+binom {13}5$$.

We then add back the cases in which all the people come from one profession. Thus we add back $$binom 65+binom 75$$

Thus the answer is $$binom {17}5-left(binom {11}5+binom {10}5+binom {13}5right)+left(binom 65+binom 75right)=boxed {4214}$$

I think it is easiest to use the principle of inclusion exclusion. Start with all $binom{17}5$ committees, ignoring the condition that each profession must appear. Then, for each profession, subtract the bad committees where that profession does not appear. So, subtract the $binom{13}5$ committees with no doctor, the $binom{11}5$ committees with no lawyer, and the $binom{10}5$ committees with no engineer. But then committees which are missing two particular professions have now been doubly subtracted, so these must be added back in to correct for this. For example, the $binom{7}5$ committees with no doctor or lawyer.

The result is
$$
binom{17}5-binom{13}5-binom{11}5-binom{10}5+ binom{7}5 +binom{6}5
$$

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.

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& {6choose3}{7choose1}{4choose1}= 20cdot7cdot4 = 560 \
hline
1&3&1& {6choose1}{7choose3}{4choose1}= 6cdot35cdot4 = 840 \
hline
1&1&3& {6choose1}{7choose1}{4choose3}= 6cdot7cdot4 = 168 \
hline
2&2&1& {6choose2}{7choose2}{4choose1}= 15cdot21cdot4 = 1260 \
hline
2&1&2& {6choose2}{7choose1}{4choose2}= 15cdot7cdot6 = 630 \
hline
1&2&2& {6choose1}{7choose2}{4choose2}= 6cdot21cdot6 = 756 \
hline
end{array}

Adding 6 gives a total of different committees:

$$ 560+840+168+1260+630+756 = 4214 $$