# Why does this solution work?

I have the following question:

There are 15 technicians and 11 chemists working in a research laboratory. In how many ways could they form a 5-member safety committee if the committee:

a) may be chosen in any way
b) must have exactly one technician?
c) must have exactly one chemist
d) must have exactly two chemists
e) may be all technicians or all chemists

To solve the questions, I used $\binom{n}{r}$ in some variation. For example, for (b) I did $\binom{15}{1}*\binom{11}{4}$.

From seeing a friend solve a question similar to this, I noticed that the top portion of the equation equalled to the total amount of technicians and chemists e.g. $26$. Therefore, the top part has to always add up to 26. Since the you can only select $5$ people in the committee, the bottom portion has to add up to $5$. Therefore, for each of the questions, I did exactly that. I kept the top equivalent to $26$ and the bottom equivalent to $5$.

For example, for (c) I did $\binom{15}{4} * \binom{11}{1}$ which is correct but I do not understand how this works or more importantly why this works and what the logic behind it is.

Thanks a lot for you time!

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For (c), you can see that the answer is $$\small\underbrace{(\#\text{ of ways of choosing the 1 chemist out of 11 chemists})}_{\Large\binom{11}{1}}\underbrace{(\#\text{ of ways of choosing the 4 techs out of 15 techs})}_{\Large\binom{15}{4}}$$ (the other 4 people chosen must be technicians, because the problem specifies exactly one chemist)
For (a), since we are not distinguishing between the two subgroups of people, the answer is just $$(\#\text{ of ways of choosing 5 people out of 26 people})=\binom{26}{5}$$
I believe (e) is asking all chemists or all technicians. I did $\binom{15}{5} + \binom{11}{5}$ – Jeel Shah Feb 27 '13 at 23:47
That's what I figured, but I didn't see why the question would stop using "must" :) Also, did you mean $$\binom{15}{5}+\binom{\mathbf{11}}{5}\quad?$$ – Zev Chonoles Feb 27 '13 at 23:48
Well, that's thinking of it more like a probability question. I like to think of it as a decision tree; you traverse each "path" of decisions. In (e), you have one decision to make first: do you want techs, or chemists? If you choose techs, you get $\binom{15}{5}$ choices; if you choose chemists, you get $\binom{11}{5}$ choices. In (c), it's the same reasoning, disguised as multiplication: let's say you want to choose your one chemist first. You have 11 choices. If you choose chemist A, you get $\binom{15}{4}$ choices of techs; if you choose chemist B, you get $\binom{15}{4}$ choices of techs; – Zev Chonoles Feb 28 '13 at 0:04