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.
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.