Ask your own question, for FREE!
Mathematics 15 Online
OpenStudy (anonymous):

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

OpenStudy (anonymous):

dont get this one, could someone show me the steps?

OpenStudy (lgbasallote):

"chosen" so you use combinations

OpenStudy (anonymous):

from 9 choose 4. 9C4 = ...

OpenStudy (lgbasallote):

arent you missing something?

OpenStudy (anonymous):

362880?

OpenStudy (lgbasallote):

that's 9!

OpenStudy (anonymous):

don't forget to divide by all the same combinations (n-r)!

OpenStudy (lgbasallote):

\[9C4 \implies \frac{9!}{4!(9-4)!}\]

OpenStudy (anonymous):

dividing by (n-r)! is what separates the combination formula from the permutation, because you are not counting all the different arrangements

OpenStudy (lgbasallote):

isnt dividing r! what separates it?

OpenStudy (lgbasallote):

\[nCr = \frac{n!}{r!(n-r)!}\] \[nPr = \frac{n!}{(n-r)!}\]

OpenStudy (lgbasallote):

dividing by r! is just the difference i see =))

OpenStudy (anonymous):

126!!

OpenStudy (lgbasallote):

i hope those ! arent factorials :p haha

OpenStudy (lgbasallote):

anyway yeah. 9C4 = 126

OpenStudy (anonymous):

ya:P

OpenStudy (anonymous):

Ooops, I meant r! not (n-r)! for all the above posts. Sorry :o

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!