Have the following problem to solve, is this the correct method, how many ways can a subcommittee of 5 be selected from a committee of 12, so,, 12C5 = 12!/7!.5! =479001600/(5040)x(120) is this right?
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Hi Originally Posted by offahengaway and chips 12C5 = 792 is this right? Yes, it is
so if i had to see how many 11 letter combinations out of 17 letters could be made would that look like this 17C11 = 17!/6!.11!
Originally Posted by offahengaway and chips so if i had to see how many 11 letter combinations out of 17 letters could be made would that look like this 17C11 = 17!/6!.11! Yes & No. Yes if you talk about 17 different letters There ain't 17C11 combinations if you have 12x X and 5x Y So it depends on your letters you are talking about.
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