How is multiplication in a counting subsets problem justified? Consider a set of $12$ people: $5$ men and $7$ women. To count all the $5$ people teams consisting of $3$ men and $2$ women, we choose $3$  men out of $5$ and $2$ women from $7$: $ {5 \choose 3} {7 \choose 2}$. Why do we use multiplication here and how can we justify it? 
 A: Imagine writing all the possible teams on slips of paper; we want to count how many slips of paper this uses.  Now imagine organizing these slips of paper into stacks, putting two slips in the same stack if the teams written on the two slips have the same men. So each stack has $\binom72$ slips in it, one for each choice of women.  There are $\binom53$ stacks, one for each choice of men.  Finally, $\binom53$ stacks of $\binom72$ slips each makes $\binom53\binom72$ slips altogether.
A: Lets look to a simpler problem. 
Suppose I have two sheets of paper, a blue and a yellow. I can write an A, a B, a C or a D on the blue paper, and I can write an A, a B or a C on the yellow. How many possibilities are there?


*

*If I write an A on the blue paper, I have three possibilities for the
yellow paper.

*If I write a B on the blue paper, I have three possibilities for the
yellow paper.

*If I write a C on the blue paper, I have three possibilities for the
yellow paper.

*If I write a D on the blue paper, I have three possibilities for the
yellow paper.


Therefore I have for every possibility of the blue paper three possibilities for the yellow.
Since I have four possibilities of the blue paper, I have 12 possibilities in total. 
You case is similiar, but the above one is easier to grasp. 
A: Lets make the numbers less and then think why do we multiply:
Now Lets say we have 6 people 
3 men,3 women.
I need to make a team of 2 :
1 men, 1 women
Now if we go by the standard procedure since order doesn't matter, we will be doing
$$\binom{3}{1}\binom{3}{1} = 9$$
Now if we add equals 6
Lets now see which is right...
Let the men be $A,B,C$ and the women be $Q,R,S$. We can make a team with A and the rest of the women and that gives me 3 choices,now same with B which also gives me 3 choices. So adding up gives me 6 choices. But wait!!
$C$ is still left . So with $C$ it gives me total 9 choices.
So what we see that when we add up the choices we do miss some choices which are present.
