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its on permutations and combinations algebra 2

It is combinations. And the result is same for the all four dinners.

how do i do it?

I can do the problems that are already set up but not word problems where i have to set it up

brb

eek i got a weird answer

what did you get?

\[11/18144\]

that was evaluating the right side

i got 220

is that for the first dinner

yes

ok how do i find the others

not sure how to solve that?

its 9/3??

Check my answer, I generalized the formula for you

\[\left(\begin{matrix}c \\ n\end{matrix}\right) = \frac{c!}{n! \times (c-n)!}\]

i got 84

yeah, probably, I didn't calculated

ok 3rd?

How many items do we left? We used 6 of them so far right?

mm let me see ill try

i got 20

yes. I think

ok for the fourth then all of the variables in the formula you gave are 3 right

yes, you have learned ;)

i think so but the problem is that I'm getting 1/6?

you should get 1. Check if you didn't make any calculation mistakes.

or i guess i would just say one because there isn't anything left

yes :)

oh haha okay thank you so much (:

Anytime ;)