Arrangement of $15$ letters with restrictions on which letters are permitted in which positions How many $15$ letter arrangements of $5$ As, $5$ Bs and $5$ Cs have no As in the first $5$ positions, no Bs in the next $5$ positions, and no Cs in the last $5$ positions?
I am not able to frame the equation. I presumed that we fix the first $5$ positions if arrangement is $10 \times 9 \times 8 \times 7 \times 6 \times 5$
I am not able to arrange the middle $5$ positions as some element of C must be overlapping.
 A: If we choose which of the middle five positions are occupied by $A$s, the others must be occupied by $C$s (since no $B$s are permitted in the middle five positions).  If we then choose which of the last five positions are occupied by the remaining $A$s, the remainder must be occupied by $B$s (since no $C$s are permitted in the last five positions).  If we then choose which of the first five positions are occupied by the remaining $B$s, the remainder must be occupied by $C$s.
If $k$ of the middle five positions are occupied by $A$s (with the remaining $5 - k$ positions occupied by $C$s), $5 - k$ of the last five positions are occupied by $A$s with the remaining $k$ occupied by $B$s, which means $5 - k$ of the first five positions are occupied by $B$s (and the remaining $k$ are occupied by $C$s).  Since choosing which of the middle five positions are occupied by $A$s, which of the last five positions are occupied by $A$s, and which of the first five positions are occupied by $B$s completely determines the sequence, the number of admissible arrangements is 
$$\sum_{k = 0}^{5} \binom{5}{k}\binom{5}{5 - k}\binom{5}{5 - k} = \binom{5}{0}\binom{5}{5}^2 + \binom{5}{1}\binom{5}{4}^2 + \binom{5}{2}\binom{5}{3}^2 + \binom{5}{3}\binom{5}{2}^2 + \binom{5}{4}\binom{5}{1}^2 + \binom{5}{5}\binom{5}{0}^2$$
