Question

How many words can we build using exactly 4 A's, 4 B's and 4 C's if the first 4 letters cannot be A's, the second 4 letters cannot be B's and the third 4 letters cannot be C's?

Answer 1

Assume that exactly $N$ A’s stand on the last 4 positions (so $𝑁∈\{0,1,2,3,4\}$ ). It follows that there are $4−𝑁$ B’s on the last 4 positions and $4−𝑁$ A’s on the middle 4 positions. Also it follows that there are $𝑁$ C’s on the middle 4 positions and $4-𝑁$ C’s on the first 4 positions.

So there are $\begin{pmatrix} 4 \\N \end{pmatrix}$ ways to place $N$ A’s on the last 4 positions. Then there are $4-N$ ways to place the rest A’s on the middle 4 positions. Finally, there are $\begin{pmatrix} 4 \\N \end{pmatrix}$ ways to place $N$ B’s on the first 4 positions. So if we place $N$ A’s on the last 4 positions, $4-N$ A’s on the middle 4 positions and $N$ B’s on the first 4 positions, then the rest positions can be filled in the only way –C’s on the first and on the middle 4 positions, B’s on the last 4 positions.

Hence the number of words with $N$ A’s on the last 4 positions is equal to

$\begin{pmatrix} 4 \\N \end{pmatrix} \cdot \begin{pmatrix} 4 \\4-N \end{pmatrix} \cdot \begin{pmatrix} 4 \\N \end{pmatrix} = \begin{pmatrix} 4 \\N \end{pmatrix}^3$

and the total number is

$\displaystyle \sum_{N=0}^4 \begin{pmatrix} 4 \\N \end{pmatrix}^3 = 1^3 + 4^3 + 6^3+4^3+1^3=1+64+216+64+1=346$

Answer: 346