The problem is:
How many words of $12$ letters can I get if the word includes $4$ letters of C and $5$ letters of D and on the other spots of the word can be any letter of the $26$ letters of alphabet?
The word does not need to make sense; it is just a string. I have solved the exercise as: $12$ spots are available for $4$ Cs is $\frac{12!}{4!8!}$. $8$ spots available for $5$ Ds is $\frac{8!}{5!3!}$. So we have: $\frac{12!}{4!8!} \cdot \frac{8!}{5!3!}$. We have also $3$ letters not specific of the alphabet which has $26 - 2$ letters already used $= 24$ letters. So from the rule of multiply in statics we have $24 \cdot 24 \cdot 24 = 24^3$.
So my question is: we have $\frac{12!}{4!8!} \cdot \frac{8!}{5!3!} \cdot 24^3$ different words or $\frac{12!}{4!8!} \cdot \frac{8!}{5!3!} + 24^3$ words or we calculate it with a totally different method? Thanks