How many ways can we put $5$ red balls, $4$ green balls and $3$ white balls into $12$ slots?

Question

Would it be $12!$ or $\dfrac{12!}{5!4!3!}$? I'm confused here.

It could depend on whether for example swapping the order of two red balls would count as a different arrangement or not. — Henry, Sep 23 '14 at 06:43
@Henry I see, but nothing like that is mentioned in the question. So I'm guessing the question is more or less .. how many ways to put 12 balls in 12 slots? — sparta93, Sep 23 '14 at 06:47
I think it's more likely that the red balls are to be considered indistinguishable, likewise the green, likewise the white. — Gerry Myerson, Sep 23 '14 at 07:01
@Henry they wouldn't bother with the colors if red balls were distinguihsable from each other. — MT_, Sep 23 '14 at 07:13
@Soke: You may be correct, if you assume that questions only ever give useful information. — Henry, Sep 23 '14 at 07:28
see $N_{all}$ in this answer. According to it it should be 12!/5!4!3! — Freddy, Sep 23 '14 at 09:06

score 2 · Accepted Answer · answered Sep 23 '14 at 07:14

2

Since (red/green/white) balls are indistinguishable from other (red/green/white) balls, the answer is $\dfrac{12!}{5!4!3!}$, not $12!$.

answered Sep 23 '14 at 07:14

MT_

1 Answers1