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The question says:in how many ways we could put 4 black,4 white,4 red balls in 6 different boxes?

boxes are distinguishable,black balls are identical,red balls are identical,and white balls are identical.

one or more box can remain empty.

i know number of ways to put k indistinguishable balls to n distinguishable boxes.but i don't know how to attack this problem because 3 different kind of balls.

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  • $\begingroup$ Each color is independent from each other. Just calculate the possibilities for one color, and then multiply by each other color to get the total possibilities. $\endgroup$ – Asimov Jul 24 '14 at 4:49
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Hint: Because colors are distinguishable, the answer is just how you put 4 red balls in 6 boxes times 4 white balls in 6 boxes, times 4 black balls in 6 boxes. Each color has the same number of balls, so they have the same number of combinations, so the answer is the number of ways to put 4 balls in 6 boxes, cubed (one time for each color, so three times multiplied)

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I will write out the approach that @Asimov posted.

Let's consider the 4 red balls. I can put red ball #1 into any of the six boxes. I can also put red ball #2 into any of the six boxes. And so on for red ball #3 and red ball #4. Therefore, I the 4 red balls into 6 boxes in 6^4 different ways.

Since, I also have 4 white and 4 black balls, I also have 6^4 possibilities for each of those.

This brings the total number of ways for all three to (6^4) * (6^4) * (6^4) or 6^12.

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