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Given $4$ blanks, find the number of ways you can fill four Reds, one Blue, one Green and​ one Yellow so that none of the blanks is filled up with $4$ reds and none of the blanks remain empty.

I cannot get the final answer. All I can do is find the number of ways four blanks can be filled using the colours but then the permutation in each case is something which I cant do. Here's what I did : add up the coefficient of $x^4, \cdots, x^7$ in $(x^4+\cdots x^0)(x^1+x^0)^3$. But that doesn't help.

Mathejunior
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  • Huh, so? What to do next, then? – Mathejunior Feb 26 '18 at 16:23
  • Can I put multiple colors on a same place or just put three extra colors out of this calculation? – user061703 Feb 26 '18 at 16:34
  • @TrầnThúcMinhTrí you can put multiple colours at the same place. But just ensure that you can't have all 4 A's at the same place, else, anything is fine as long as there's ≥ 1 colour at every location – Mathejunior Feb 26 '18 at 16:36
  • I am imagining the blanks as distinguishable boxes; but within each box, order is unimportant. So for instance (RRG)(RB)(R)(Y) is the same as (RGR)(RB)(R)(Y); but (RRG)(RB)(Y)(R) is different. Is that the correct interpretation? – paw88789 Feb 26 '18 at 16:41
  • @paw88789 it's a correct interpretion. Btw, it's not necessary that all 7 colours have to be used up. Like, you can also take (RR)(Y)(B)(R) as well or anything you wish – Mathejunior Feb 26 '18 at 16:44

1 Answers1

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I would just count by hand, starting with the choices for the red. The reds can be distributed $(3,1,0,0), (2,2,0,0), (2,1,1,0), (1,1,1,1)$. Now the two zeros have to get one of the other colors, so you get $(3X,1,Y,Z), (3,1X,Y,Z), (3,1,XY,Z)$ as possibilities. There are six ways to assign colors to $XYZ$ and each can be assigned to the boxes in $12$ ways except $(3,1,XY,Z)$ is $24$ because it does not have the two single colors to swap, but assigning $Y$ and $Z$ can be swapped so we have $2\cdot 6 \cdot 12+ 6 \cdot 12=216$. Keep going with the rest.

Ross Millikan
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  • How's it $6\cdot 24$? In that case, you are considering $(3,1,XY,Z)$ and $(3,1,YX,Z)$ as two different cases. But they are rather similar – Mathejunior Feb 27 '18 at 05:54
  • You are correct. The counting is hard and I missed that. I will fix. – Ross Millikan Feb 27 '18 at 06:06
  • sure, that's not an issue. Thanks for your help (always) – Mathejunior Feb 27 '18 at 06:06
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    I have fixed it and believe I have this case correct. I made some mistakes before the original post, so it takes a lot of care. – Ross Millikan Feb 27 '18 at 06:08
  • Sir, sorry to interrupt but I think that it should be $2\cdot 3! \cdot 4!$ instead of $2\cdot 6 \cdot 12$ because the four places can be arranged in $24$ ways. Also, the last case will be like $4!\cdot 3$. So, I think that the answer should be $4!3!+4!3!+4!3=360$ if I am not wrong. – Mathejunior Feb 27 '18 at 06:37
  • Sir, sorry to interrupt but I think that it should be $2\cdot 3! \cdot 4!$ instead of $2\cdot 6 \cdot 12$ because the four places can be arranged in $24$ ways. Also, the last case will be like $4!\cdot 3$. So, I think that the answer should be $4!3!+4!3!+4!3=360$ if I am not wrong. – Mathejunior Feb 27 '18 at 06:39
  • For the first two patterns arranging the four places in $4!$ ways double counts because you could assign the $Y$ and $Z$ colors in reverse, then swap those two places. One or the other of the $3!$ and $4!$ has to be divided by $2$ to account for this. The last case is similar but it is the $X$ and $Y$ that can be swapped. – Ross Millikan Feb 27 '18 at 15:28
  • Yeah, that. We have to divide by $2$. – Mathejunior Feb 27 '18 at 15:30
  • So, $(2X,2,Y,Z)$ can be done in $\frac{4!\cdot 3!}{2!}$ ways, am I correct? – Mathejunior Feb 27 '18 at 15:33