TITLE: How many ways can you choose team of 5 people out of 7 men and 6 women in which there are at least 3 men?
QUESTION [5 upvotes]: I am confused by this question.
I solved it by selecting 3 men first out of 7 men and then selecting 2 people out of 10 remaining person ( 4 men and 6 women ) . 
So my answer is C(7,3) * C(10,2)  = 1575 .
But my friend solved it by 
selecting 3 men out of 7 men and selecting 2 women out of 6 (first case)  + 
selecting 4 men out of 7 men and selecting 1 women out of 6 (second case) + 
selecting 5 men out of 7 men and selecting 0 women out of 6 (third case).
So his answer is C(7,3) * C(6,2) + C(7,4) * C(6,1) + C(7,5) * C(6,0) = 756.
I would like to know who is right and why.

REPLY [2 votes]: The right way is to split it into disjoint events and then sum up their amounts:

Exactly $\color\red3$ men: $\binom{7}{\color\red3}\cdot\binom{6}{5-\color\red3}$
Exactly $\color\red4$ men: $\binom{7}{\color\red4}\cdot\binom{6}{5-\color\red4}$
Exactly $\color\red5$ men: $\binom{7}{\color\red5}\cdot\binom{6}{5-\color\red5}$


The total amount is therefore:
$$\sum\limits_{n=3}^{5}\binom{7}{n}\cdot\binom{6}{5-n}=756$$