This question has been asked here: find the number of rearrangements of 11223344 that contain no two consecutive equal digits
My question is slightly different. In the linked thread they use multinomial coefficients, I am trying to do the same thing using the binomials.
Suppose we have eight boxes labeled $A$ through $H.$ Also assume we have eight balls labeled $1, 1, 2, 2, 3, 3, 4, 4.$
There are seven ways to place two balls labeled $1, 1$ into the adjacent boxes. These possibilities are $AB, BC, CD, DE, EF, FG, GH.$ For each of these possibilities there are $\binom62\binom42$ ways to place the rest of the balls. We must repeat this process for the pairs $2, 2; 3, 3; 4, 4$. In total there are $7 \cdot \binom62\binom42 \cdot 4$ ways to count $\sum_{i = 1}^4|S_i|$ where $S_i$ is a set of words where integers $ii$ are consequtive.
Now suppose we want to place the pairs $1, 1$ and $2, 2$ into the seven adjacent box pairs. My first guess was to choose a box pair from those listed above. There are seven possibilities. Then choose another box pair from the remaining six. Altogether there are $7 \cdot 6 \cdot \binom42 \cdot 6$ ways. But this is wrong. The correct answer is $6 \cdot 5 \cdot \binom42 \cdot 6.$ Looks like we are choosing the first box pair out of six possible pairs, not seven. Why is that? One problem seems to be that if we choose the boxes $CD,$ then we can't choose $BC$ and $DE$ leaving only four possible pairs whereas if we choose $AB$, we have five possible pairs left. This probably complicates things. Can someone, please, explain how to think correctly to obtain $6 \cdot 5 \cdot \binom42 \cdot 6$ rather than $7 \cdot 6 \cdot \binom42 \cdot 6.$
