Counting rules and probability A committee of 6 members is chosen from 7 engineers, 5 chemist and a doctor so as to consist at least 2 chemist, a doctor and at least an engineer but if one particular engineer and one particular chemist refuse to serve together on the same committee​, how many ways can this be done?
 A: First we can observe that we are technically selecting 5 people from 5 chemists and 7 engineers as we have to have that one doctor no matter what. We can first count the total possible ways to select the remaining 5 people, then subtract the number of times that one pair of chemist and engineer who refuse to work together.
The number of ways to select 5 people is $$5*4*7*9*8$$
This is because we must first select at least two chemist, hence $5*4$, then select at least one engineer, hence the $7$. Then we are free to choose the remaining 9 for the last two spots, hence $9*8$
The number of ways where that one single bad pair occurs is $$1*1*4*9*8$$
This is because we are imposing the restriction of that one particular pair having to be selected, hence $1*1$. From here we see we already have one chemist and one engineer, hence we must select just one more chemist from the 4 remaining , hence the $4$. Then we are free to select the remaining 9, hence the $9*8$.
Knowing the total possibilities of selecting 5 such that there's at least two chemist and at least one engineer, and the possibilities that have that one pair that won't work, we can just subtract those two. Thus
$$5*4*7*9*8-1*1*4*9*8=9792$$
Our answer is 9792 total ways to select the 5 people
