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? 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. 
 A: 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:
$$\sum\limits_{n=3}^{5}\binom{7}{n}\cdot\binom{6}{5-n}=756$$
A: Extended comment:
I agree with @drhab. Also, because this is under a (probability) tag, I wanted to mention that this kind of computation is not
just a drill. It is the basis of the hypergeometric distribution.
First, one can use R, or other statistical software, to verify your
friend's numerical answer. (Software is especially convenient in such
computations when the numbers get larger: for example, choosing
at least 10 men when drawing 15 from among 20 men and 30 women.)
 i = 3:5
 choose(7,i)*choose(6, 5-i)
 ## 525 210  21
 sum(choose(7,i)*choose(6, 5-i))
 ## 756

Now for the hyperometric distribution. Suppose we have an urn
with 13 chips, of which 7 are blue and 6 are pink. If we
withdraw 5 chipss at random (without replacing a chip once
it is drawn), what is the probability we will get at least 3
blue chips among the 5 drawn?
There are $C(13, 5) = {13 \choose 5}$ equally likely ways
in which to choose 5 chips from among 13. Your friend's
answer is the number of ways to make the choice so that we
get at least 3 blue chips. Let $X$ be the number of
blue chips drawn in such an experiment. Then
$$P(X \ge 3) = \sum_{i=3}^5 P(X = i) = \frac{\sum_{i=3}^5 {7 \choose i}{6 \choose 5-i}}{{13 \choose 5}} =\frac{756}{1287} =  0.5874.$$
In R, this computation is
 dhyper(i, 7, 6, 5)
 ## 0.40792541 0.16317016 0.01631702
 sum(dhyper(i, 7, 6, 5))
 ## 0.5874126
 sum(dhyper(i, 7, 6, 5))*choose(13,5)
 ## 756

Below is a graph of this entire hypergeometric distribution
with values of $X$ running from $0$ through $5.$ The three bars
corresponding to $P(X \ge 3)$ are colored dark blue. Their combined
height is $0.5874.$ 
All six of the values in this distribution (the combined height of
all six bars) must sum to $1,$ because
${7\choose 0}{6 \choose 5}+{7 \choose 1}{6 \choose 4} + \cdots
+{7 \choose 5}{6 \choose 0} = {13 \choose 6}.$

