How many words can we build using exactly 5 A's, 5 B's and 5 C's How many words can we build using exactly 5 A's, 5 B's and 5 C's if the first 5 letters cannot be A's, the second 5 letters cannot be B's and the third 5 letters cannot be C's? 
Hint: Group the different ways according to the number of B's in the first group. 
 A: Split up in words $w_1$, $w_2$ and $w_3$ consisting each of $5$ letters. The final word of $15$ letters is concatenation $w_1w_2w_3$.
Let $k\in\{0,1,2,3,4,5\}$ denote the number of letters $B$ that are used in $w_1$. 
Then there are $5-k$ letters $C$ in $w_1$. 
The remaining $k$ letters $C$ must be used in $w_2$. Next to them $5-k$ letters $A$ must be used in $w_2$. 
Finally $5-k$ letters $B$ and $k$ letters $A$ remain to be used in $w_3$.
For $w_1$ (and also for $w_2$ and $w_3$) there are $\binom5{k}$ possibilities so for the whole word there are $\binom5{k}^3$ possibilities.
So the final answer is:$$\sum_{k=0}^5\binom5{k}^3$$
A: A picture is said to be worth a thousand words, so here is a colorful intuitive explanation.
Start with
$\color{red}{B B B BB}||\color{green}{ C C C C C}||\color{blue}{A A  A A A}$, then 
$\color{red}{B B B B}\color{green}{ C||C C C C}\color{blue}{A|| A  A A A}\color{red}{B}$ , then
$\color{red}{B B B}\color{green}{ C C || C C C}\color{blue}{A A || A A A}\color{red}{B B}$, etc
Notice the symmetry involved, and that you move from one to the next by simply shifting letters by one to the left (in a circle)
Can you compute $\binom{5}{0}^3 + \binom{5}{1})^3 + ..... + \binom{5}{5}^3$ 
