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anonymous
 3 years ago
In how many ways can a gymnastics team of 4 be chosen from 9 gymnasts?
anonymous
 3 years ago
In how many ways can a gymnastics team of 4 be chosen from 9 gymnasts?

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dont get this one, could someone show me the steps?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0"chosen" so you use combinations

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0from 9 choose 4. 9C4 = ...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0arent you missing something?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0don't forget to divide by all the same combinations (nr)!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[9C4 \implies \frac{9!}{4!(94)!}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dividing by (nr)! is what separates the combination formula from the permutation, because you are not counting all the different arrangements

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0isnt dividing r! what separates it?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[nCr = \frac{n!}{r!(nr)!}\] \[nPr = \frac{n!}{(nr)!}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dividing by r! is just the difference i see =))

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i hope those ! arent factorials :p haha

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0anyway yeah. 9C4 = 126

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Ooops, I meant r! not (nr)! for all the above posts. Sorry :o
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