How many $6$-letter words can be formed from (A,B,C,C,D,E,F,G), where the Cs cant be near each other and the other letters can be the same?

Question

How many $6$-letter words (meaning doesn't matter) can be formed from (A,B,C,C,D,E,F,G), where the Cs cant be near each other and the other letters can be the same? Also there must be $2$ Cs in each of them. I tried to solve it like this, but I'm pretty sure I'm wrong.

Total words$= 7!$

Words with at least one $C=6!$

Words where both Cs are near each other$= 6! \times 2!$

Words with 2 Cs in them and not near each other: $(7!-6!)-6!\times 2!$

Thank you in advance

Answer 1

Strategy: Based on the rules explained in the comments that each letter may appear only as often as it appears in the multiset $\{A, B, 2 \cdot C, D, E, F, G\}$, that the six-letter word must contain exactly two Cs, and that the two Cs cannot be adjacent:

1. Choose four letters from among the six distinct letters A, B, D, E, F, G.
2. Arrange the four distinct letters you have chosen in a row.
3. Doing so creates five spaces in which to place a C. $$\square L_1 \square L_2 \square L_3 \square L_4$$ where $L_i$, $1 \leq i \leq 4$, is the $i$th letter of the arrangement of the four distinct letters you have chosen.
4. To ensure that the two Cs are not adjacent, choose two of these five spaces in which to place a C.