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In how many ways can a gymnastics team of 4 be chosen from 9 gymnasts?
 one year ago
 one year ago
In how many ways can a gymnastics team of 4 be chosen from 9 gymnasts?
 one year ago
 one year ago

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mynameisjanelleBest ResponseYou've already chosen the best response.0
dont get this one, could someone show me the steps?
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.2
"chosen" so you use combinations
 one year ago

apple_piBest ResponseYou've already chosen the best response.1
from 9 choose 4. 9C4 = ...
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.2
arent you missing something?
 one year ago

apple_piBest ResponseYou've already chosen the best response.1
don't forget to divide by all the same combinations (nr)!
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.2
\[9C4 \implies \frac{9!}{4!(94)!}\]
 one year ago

apple_piBest ResponseYou've already chosen the best response.1
dividing by (nr)! is what separates the combination formula from the permutation, because you are not counting all the different arrangements
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.2
isnt dividing r! what separates it?
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.2
\[nCr = \frac{n!}{r!(nr)!}\] \[nPr = \frac{n!}{(nr)!}\]
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.2
dividing by r! is just the difference i see =))
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.2
i hope those ! arent factorials :p haha
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.2
anyway yeah. 9C4 = 126
 one year ago

apple_piBest ResponseYou've already chosen the best response.1
Ooops, I meant r! not (nr)! for all the above posts. Sorry :o
 one year ago
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