A company has 7 designers, 14 manufacturers, 4 testers, 5 in sales, and two in accounting. A committee of 6 people is to be formed.
How many different committees can be formed if there must be exactly 2 from manufacturing?
I know that first step should be to choose the 2 from manufacturing. So that would be 14 choose 2.
What should I do from here? I'm fairly new to Combinatorics and I haven't quite got the thinking of it yet.
On
There are $18$ people left to choose from, and we need to fill an additional four seats in the committee. That's $\binom{18}{4}$. Now multiply the two and you have your answer.
If we change to at least two from accounting, then we can do it like this: (each one is calculated exactly the same way as the first part)
and then add them. If that seems a bit long, then there is a simpler way: Instead count the number of committees that do not fulfill the criteria, count the total number of committees without any criteria and subtract one from the other.
Committees that do not fulfill the criteria are committees where either
and again, you calculate each of these exactly as you did the first part.
Unfortunately, there is no easy way to calculate this. What I mean is that if the numbers had been larger so that whichever of the two approaches here you chose you would get a large number of cases, then you really do have to calculate a large number of cases.
Yes you need to choose two from manufacturing $\binom{14}{2}$ and then 4 from the remaing 18 people that is $\binom{18}{4}$, thus
$$N=\binom{14}{2}\binom{18}{4}=\frac{14!}{2!12!}\frac{18!}{4!14!}=278'460$$
committee of 6 people can be formed.