Arrangements of ABCDDDEEEFFF where at least two Ds are adjacent and at least two Es are adjacent.

Question

How many arrangements are possible from a word ABCDDDEEEFFF where at least two Ds must be adjacent and at least two Es must be adjacent.

Number of arrangements where at at least two Ds are adjacent
$=\dfrac{12!}{3!.3!.3!}-\dfrac{9!}{3!.3!}\times\dbinom{10}{3}$

Number of arrangements where at at least two Es are adjacent
$=\dfrac{12!}{3!.3!.3!}-\dfrac{9!}{3!.3!}\times\dbinom{10}{3}$

But, unable to apply both the constraints together. please help how we can find the count when (at least two Ds are adjacent) and (at least two Es are adjacent)

Answer 1

Let's replace two of the $D$s with $X$, and two of the $E$s with $Y$. So the arrangements of $ABCXDYEFFF$ are simply $\dfrac{10!}{3!}$.

However we have overcounted slightly, because $XD \equiv DX$ and $YE \equiv EY$. So we need to remove the $\dfrac{9!}{3!}$ cases where each of these is overcounted then add back in the $\dfrac{8!}{3!}$ cases where both are happening and we have removed the overcount twice.

So $$\frac{10!}{3!} - 2\frac{9!}{3!}+ \frac{8!}{3!}= 604800 - 2\cdot 60480 + 6720 = 490560$$

Answer 2

The number of ways for the 2 leftmost D's to be adjacent and the 2 leftmost E's to be adjacent can be computed by thinking of those D's as a single D, and likewise for E:

$$\frac{10!}{2! \cdot 2! \cdot 3!}$$

The same is true for leftmost D's and rightmost E's, etc. We can now find the total count using inclusion-exclusion. For notation, let $D_{L}$ and $D_R$ be the sets where leftmost/rightmost D's are adjacent, and likewise for $E_L$ and $E_R$. So we're trying to find:

$$\#[(D_L \cap E_L) \cup (D_L \cap E_R) \cup (D_R \cap E_L) \cup (D_R \cap E_R)]$$

Pairwise intersections: $(D_L \cap E_L) \cap (D_R \cap E_L) = (D_L \cap D_R) \cap E_L$. So the leftmost E's are adjacent, and all $3$ D's are adjacent, so we can treat them as a single letter to get:

$$\#((D_L \cap D_R) \cap E_L) = \frac{9!}{2! \cdot 3!}$$

The same is true for 4 of the 6 pairwise intersections. The other two are $(D_L \cap E_L) \cap (D_R \cap E_R)$ and $(D_L \cap E_R) \cap (D_R \cap E_L)$, which require all three D's and E's to be adjacent. We get:

$$\#((D_L \cap D_R) \cup (E_L \cap E_R)) = \frac{8!}{3!}$$

Triple/Quadruple intersections: All of these result in $(D_L \cap D_R) \cap (E_L \cap E_R)$. So the final answer is:

$$4 \cdot \frac{10!}{2! \cdot 2! \cdot 3!} - 4\cdot \frac{9!}{2! \cdot3!} - 2 \cdot \frac{8!}{3!} + 4 \frac{8!}{3!} - \frac{8!}{3!}$$

Simplifying:

$$= \frac{8!}{3!}\left(90 - 18 - 2 + 4 - 1\right) = \frac{73 \cdot 8!}{3!} = 490560$$