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There is a group of people, 30 women and 33 men (total 63). A group must be randomly formed of 3 women and 2 men. How many combinations are possible?

My logic so far: combination (not permutation), order unimportant and repetition unallowed:

(30   (33
--- * --- = 4060 * 528 = 2143680 possible combinations
 3)    2)

I'm not sure about the repetition, should it be allowed? It would change the result.

user395040
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  • This page will teach you how to type in MathJax and $\LaTeX$ on this site so you can write things like $\binom{30}{3}$ without having to rely (poorly) on using ascii to do so. – JMoravitz Dec 01 '16 at 21:47
  • The calculation is valid. – Peter Dec 01 '16 at 21:49
  • With repetition I mean that the formula would become (r+n-1)!/r!(n-1)! as opposed to n!/r!(n-r)! – user395040 Dec 01 '16 at 21:51
  • We clearly have no repetition here. Every person can be chosen only once. At least, this would be my interpretation of the text. – Peter Dec 01 '16 at 21:57
  • @user395040 but would the "outcome" (Alice, Alice, Alice, Bob, Charlie) count as "a group of 3 women and 2 men" (where all three Alices refer to the same person)... that sounds to me like a single woman and two men, not three women and two men... – JMoravitz Dec 01 '16 at 21:57

1 Answers1

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Pick three women ($_{30}C_3$) and two men ($_{33}C_2$). The total number of groups is then the product of these two.

In this case, you don't have repetition, unless cloning is an option. (But then you really run the risk of groupthink!)

John
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