Three men (out of 7) and three women (out of 6) will be chosen to serve on a 7 member committee. In how many ways can the committee be formed?
I did 7C3 to get 35 men.
Then i did 6C3 to get 20 women.
Then i decide to add up 20 + 35 and get 55 but it is suggested i have to multiply 35 and 20 instead. I want to know why is it that we are multiplying 35 and 20 instead of adding them up?
On
You could also think about it this way. It doesn't matter the order you choose men and women, so you might as well choose all the men that you want on the committee then all the women. There are 7 total spots, _ _ _ _ _ _ _. Let's fill the first spot with a man. How many ways can we do this--7. So we have 7 _ _ _ _ _ _. Let's fill the next spot with a man, there are 6 left. Continuing, this gives 7 6 5 _ _ _ _. Now we fill in the women in the same way, This gives a total of $(7\times6\times5)\times(6\times5\times4\times3)$ committees. But you should recognize this as $\binom{7}{3} \binom{6}{3}$.
In general, just think about the choices you want. I want this guy out of the 7 here AND this guy out of the 6 left here AND ..... AND this final women out of the three left here. 'AND' logic means multiplication in probability. So whenever you have a situation where you can find yourself able to make statements like that you're most likely going to have some sort of multiplication involved.
The choices of men and women are made independently. Independent $\Longleftrightarrow$ Multiply.
Why? A concrete example might help.
I want to choose 1 fruit from $\{apple,orange,banana\}$ and 1 drink from $\{water,tea\}$. Then I have $3 \times 2 = 6$ choices: $(apple,water)$, $(orange,water)$, $(banana,water)$, $(apple,tea)$, $(orange,tea)$, and $(banana,tea)$.
For each fixed choice of the first item I can choose any of the second item. So if I have $n$ choices for item #1 and $m$ choices for item #2: I get $\underbrace{m+m+\cdots+m}_{n-\mbox{times}} = n \cdot m$ choices.