'tis maybe is an indistinguishable permutation

Quote:

Originally Posted by

**jamesinsc** I thought this would be a simple problem, but I can still not come up with the correct answer.

The problem is:

12 Students are in a class. Five can go to room A, Four to room B, and Three to room C. How many ways can this happen?

Since order is not important it is a combination problem and not a permuntation problem.

So by using nCr for {5,4} = 5 and {5,3} = 10 gives me a total of fifty combinations(using the fundamental counting principal).

Would I not just figure the combination of how many students can go into room B? nCr{4,1} = 4 and then multiply 5 x 10 x 4 ?

I get 200 for the answer but according to the quiz this is not the answer.

Could someone tell me what I am doing wrong?

Thanks in advance

James

(Sleepy).........'TIS COULD BE AN INDISTINGUISHABLE PERMUTATION

use the formula: n!/p!q!r!...

12!/5!4!3!

=479001600/17280

=27,720

MAYBE YOU COULD TRY THIS ONE...(Speechless)