How many $10$ character words formed from the alphabet $\{a, b, c\}$ have exactly six $a$s? You have $\{a,b,c\}$ as your character set.
You need to create words having $10$ characters that must be chosen from the character set. Out of all the possible arrangements, how many words have exactly $6$ $a$s?
My logic:
We need $6$ $a$s in our words. The rest $4$ words must come from $\{b,c\}$.
Final answer $= \dfrac{2C1 \cdot 4 \cdot 10!}{6!}$
The $6!$ comes from the $6$ duplicate $a$s in our words.
Is my logic fine?

Since you need exactly $6$ copies of "a" in the ten characters, then there are ${}_{10}C_6$ possible ways to choose where they will go. For each of the remaining four characters, there are exactly two ways to choose, meaning there are $2^4=16$ ways to choose the remaining characters. Thus, we have $${}_{10}C_6\cdot 16=\frac{10!\cdot 16}{4!\cdot 6!}=\frac{10\cdot 9\cdot 8\cdot 7\cdot 16}{4\cdot 3\cdot 2}=10\cdot 3\cdot 7\cdot 16=3360$$ possible words.
Your number comes to $$2\cdot 4\cdot 10\cdot 9\cdot 8\cdot 7=10\cdot3\cdot 7\cdot 16\cdot 12=3360\cdot 12=40320.$$ To see why this isn't reasonable, note that there are only $3^{10}=59049$ possible words that can be constructed with ten characters from that set, so you're claiming that the great majority of them have exactly $6$ copies of "a." But by the same reasoning, we would also have the great majority with exactly $6$ copies of "b," which can't also be true since there are only ten characters to go around.
