Thank you for the opportunity to help you with your question!
possible ways of choosing 3 men from 5 is: 5!/(3!2!) = 120/12 = 10
possible ways of choosing 2 women from 3 is: 3!/(2!1!) = 6/2 = 3
this gives us a total of 30 allowable combinations.
@Bryan G: what you have to realize, is that of your 6 combinations of choices, 3 of the are the same. Suppose the women are Mary, Jane and Sue.
so after the first choice, we have the follwing possibilites:
M, J, S
suppose Mary was chosen: that gives us the following possibilites:
MJ or MS (that's two)
now suppose Jane was chosen, our possible possibilites after the next choice is made are:
JM or JS. but choosing Jane then Mary, gives us the same outcome as choosing Mary, then Jane. so that's a duplication, and the only new combination is:
JS (that makes 3).
now suppose Sue is chosen first. that just leaves Mary and Jane left. but SM already occured as MS, and SJ already occured as JS, so there are no new possibilities left.
the same reasoning applies to the choices for the men, your list contains duplicates.
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