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In how many ways is it possible divide 6 identical blue , 6 identical green and 6 identical red items among 3 people such that each gets equal number of items.

The problem in my book says two people but I read it wrong and started to figure it out for 3, which I couldn't. Though I know how to do the correct question I wanted to know how we would solve this for 3 people. It would be great if someone could provide me with a hint to solve this question as I don't know what approach to adopt.

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    $\begingroup$ The number you're looking for can be modeled as 3x3 squares with constrained row and column sums. It's certainly possible to calculate it with some computation, but I'm not sure that there's any sort of explicit form for it. $\endgroup$ – Steven Stadnicki Feb 28 '17 at 4:06
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    $\begingroup$ The generating function for this is $((1-xb)(1-xg)(1-xr)(1-yb)(1-yg)(1-yr)(1-zb)(1-zg)(1-zr))^{-1}$. In your case you want the coefficient of $(xyzrgb)^6$. $\endgroup$ – N. Shales Feb 28 '17 at 13:06
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    $\begingroup$ Using the following input in sage: var('x y z r g b') f(x,y,z,r,g,b)=((1-xb)*(1-xg)*(1-xr)*(1-yb)*(1-yg)*(1-yr)*(1-zb)*(1-zg)*(1-z*r))^(-1) taylor(f,(x,0),(y,0),(z,0),(r,0),(g,0),(b,0),36).coefficient(x^6).coefficient(y^6).coefficient(z^6).coefficient(r^6).coefficient(g^6).coefficient(b^6) gives a result of $\mathbf{406}$ $\endgroup$ – N. Shales Feb 28 '17 at 21:23
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    $\begingroup$ @N.Shales Okay...thank you so much $\endgroup$ – Osheen Sachdev Mar 1 '17 at 2:47

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