Question 20How many $15$-letter arrangements of $5$ A’s, $5$ B’s, and $5$ C’s have no A’s in the first $5$ letters, no B’s in the next $5$ letters, and no C’s in the last $5$ letters?$\displaystyle\sum_{k=0}^{5}\binom{5}{k}^3$$3^5 \cdot 2^5$$2^{15}$$\dfrac{15!}{(5!)^3}$$3^{15}$