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

"chosen" so you use combinations

from 9 choose 4.
9C4 = ...

arent you missing something?

362880?

that's 9!

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

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

isnt dividing r! what separates it?

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

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

126!!

i hope those ! arent factorials :p haha

anyway yeah. 9C4 = 126

ya:P

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