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From a set of $12$ colored balls, including three red, three blue, three green, and three white, $9$ balls are selected. How many ways can this be done?

I was wondering if we can use the identity $\binom nk = \binom{n}{n - k}$ to transform the given problem to an identical(?) one below

What's the number of $3$-multisets whose elements come from $\{\text{red, blue, green, white}\}$?

In other words, are the two quoted problems identical?

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Choosing 9 balls is the same as "choosing the 3 balls that aren't chosen", so your problems are identical.

Note that this is only true because you have 3 of each color ball, so you are free to take as many of each color for the "not-chosen" group as you like (since you only want three total). If the problems were instead:

From a set of 12 colored balls, including three red, three blue, three green, and three white, 8 balls are selected. How many ways can this be done?

and

What's the number of 4-multisets whose elements come from $\{\mbox{red, blue, green, white}\}?,$

they would no longer be identical problems. The second one would imply that you can choose as many red balls as you want, up to and including four, while the first would preclude choosing four red balls for the "not-chosen" group.

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