Combinatorics: A certain panel is to be composed

[NMencia09]

Hi,

This question was a data sufficiency, but I would like to know how to solve it.

A certain panel is to be composed of exactly 3 women and 2 men, chosen from 6 women and 5 men. How many different panels can be formed with these constraints?

My attempt:
look at the 2 groups individually. 3 women from a group of 6 can be chosen: 6!/3!3! = 20 ways.
two men from 5 can be chosen: 5!/2!3! ways = 10 ways.

20 ways to choose the women, 10 ways to choose the men.
20X10= 200 ways

its this last step im not quite sure about. any thoughts gmat gurus?

thanks in advance

[arnabgangully]

I think your approach is a right one

[jnelson0612]

Correct! Whenever we have two or more subgroups that are combining to make a large group, we find out the number of combinations for each subgroup and then multiply the numbers for each subgroup together. In other words, what you did. :-)