Permutations and Combinations in arranging letters

Question

Number of ways to arrange the letters $(A,A,A,A,A,B,B,B,C,C,C,D,E,E,F)$ such that no two $C's$ are together ?

My solution:

I calculated all possibilities with CC together i.e $\dfrac{14!}{5!3!2!}$ and subtracted it from total i.e $\dfrac{15!}{5!3!3!2!}$.

But my answer is wrong. Why?

Answer 1

Because the combinations where all three C are adjacent were mishandled: each case of CC C was counted separately from all the cases of C CC, when in reality they're the same. You therefore need to subtract once all different combinations that have all three C's together to get the correct number of combinations where at least two C's are together.

Answer 2

HINT: Let us denote the three $C$'s by $C_1, C_2, C_3$. There are $4$ possible ways of these $C$'s coming together: $C_1C_2, C_2C_3, C_1C_3, C_1C_2C_3$. I hope you will now be able to continue.

Answer 3

12 alphabets without C's can be arranged in 12! ways. ...(A)

We can arrange 3 C's in any of the 13 (=12+1) positions marked as * below.

$* 1 * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9 * 10 * 11 * 12 * 13 * $

(where 1, 2… 13 represent remaining alphabets)

This can be done in C(13,3) ways. ......(B)

From (A) and (B), required number of ways

$= \frac{12! × C(13,3)}{5! × 3! ×2!}$

$= \frac{12! × 13!}{10! × 3! × 5! × 3! ×2!}$