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In how many ways can you paint 90 distinct buckets, if 25 of them must be painted red, 40 of them must be painted blue, and 25 of them must be painted green?

I am right to assume that these object are mutually exclusive, if so am I right to say that the answer would be 25^90 + 40^90 + 25^90?

mvitagames
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1 Answers1

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Choose 25 of the 90 buckets to be red, 25 of the remainder to be green. The remaining 40 must automatically be blue.

$\binom{90}{25}*\binom{65}{25}=\frac{90!}{65!25!}*\frac{65!}{25!40!}=\frac{90!}{(25!)^240!}$

Sherlock Holmes
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  • was about to post a similar answer: ${90\choose 40}\cdot {50 \choose 25}$ – Rustyn Sep 21 '14 at 07:58
  • I'm not sure you did your addition correctly @ Rustyn, shouldn't that be 40 in the lower bit of the first binomial coefficient? – Sherlock Holmes Sep 21 '14 at 08:00
  • whoops should've been a 40 – Rustyn Sep 21 '14 at 08:01
  • I dont really get this, how about if we had 200 distinct bucket, 33 of them must be painted red, 66 of them must be painted blue, and 101 of them must be painted green? – mvitagames Sep 21 '14 at 08:04
  • The answer would be:

    $\binom{200}{33}*\binom{167}{66}$

    The binomial coefficient $\binom{n}{r}$ represents the number of ways that r objects may be selected from n objects. So first, we choose 33 objects from the original 200 to be red, then from the remaining 167, we choose 66 to be blue. The remaining 101 buckets must automatically be green, so there is only one choice.

    Is that a little easier to understand?

     – Sherlock Holmes Sep 21 '14 at 08:07
  • OOO, ic that's so much clear now, so basically it's like if we have 4 student and in board of director can consist of one president and 3 vice president then it the case that there is 4 way to choose a president and 1 way to choose a vice president since there is 3 vice president in a board of director. there using the product rule we get 4 way of using a boardof director – mvitagames Sep 21 '14 at 08:18