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?
 A: 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.
A: Consider the ten-character word as ten spaces to be filled. You can use the characters a, b or c to fill these spaces. We need to find the possibilities where exactly 6 spaces are filled with a's.
So, first choose 6 of these spaces to reserve them for the a's. The number of ways that can be done is 10C6. There is no further arrangement needed, because all a's are identical.
Now for the remaining 4 spaces, you have to choose between b and c. There could be 0 b's or 1 b's or 2 or 3 or 4. We add the possibilities for each case.
Case 1: (0 b) There is only one arrangement possible for this so-    1
Case 2: (1 b) The number of ways you arrange 1 b and 3 identical c's is $\frac{4!}{3!}$
You can also imagine this as choosing 3 spaces out of 4 for the c's, in which case you would do 4C3, which would again be $\frac{4!}{(3!*1!)}$
Case 3: (2 b) Similarly, arranging 2 identical c's and 2 identical b's, we get   $\frac{4!}{2!*2!}$
Case 4: (3 b) The number of ways you arrange 3 identical c's and 1 b is     $\frac{4!}{3!}$
Case 5: (4 b) Again, there's just one way you can do this. So - 1
So the final number of ways is  $$10C6 (1 + 4 + 6 + 4 + 1) $$
or $$10C6 * 16$$
I can't understand your thought process for the numerator and why you did 2C1 and multiplied it by 4 etc, so I can't really show you where your logic is faulty, but I'd do it the way I showed here.
