TITLE: Use Inclusion-Exclusion to find the number of arrangements of 4 A's, 5 B's, 6 C's, 7 D's , 8 E's with exactly 2 adjacent C's
QUESTION [6 upvotes]: I found the number of arrangements in the problem stated above by arranging the letters other than C and then counting the number of ways to insert the C's into the gaps, giving an answer of
$\displaystyle\hspace{1.2 in}\frac{24!}{4!5!7!8!}\binom{25}{5}\binom{5}{1}=281625478590456000$,
but I would like to find out how to work this problem using Inclusion-Exclusion instead.

REPLY [2 votes]: To find the number of arrangements with exactly one pair of C's, we must use a generalization of the Inclusion-Exclusion Principle.  Let $A_k$ denote the set of arrangements with $k$ pairs of adjacent C's.  
One pair of adjacent C's 
We have $29$ objects to arrange: $4$ A's, $5$ B's, $1$ CC, $4$ C's, $7$ D's, $8$ E's.  They can be arranged in 
$$|A_1| = \frac{29!}{4!5!1!4!7!8!}$$
distinguishable ways. 
Two pairs of adjacent C's 
Two disjoint pairs of C's:  We have $28$ objects to arrange:  $4$ A's, $5$ B's, $2$ CC's, $2$ C's, $7$ D's, $8$ E's.  They can be arranged in 
$$\frac{28!}{4!5!2!2!7!8!}$$
distinguishable ways.
Two overlapping pairs of C's:  We have $28$ objects to arrange:  $4$ A's, $5$ B's, $1$ CCC, $3$ C's, $7$ D's, $8$ E's.  They can be arranged in
$$\frac{28!}{4!5!1!3!7!8!}$$
distinguishable ways.
Hence, 
$$|A_2| = \frac{28!}{4!5!2!2!7!8!} + \frac{28!}{4!5!1!3!7!8!}$$
Three pairs of adjacent C's
Three disjoint pairs of adjacent C's:  We have $27$ objects to arrange:  $4$ A's, $5$ B's, $3$ CC's, $7$ D's, $8$ E's.  They can be arranged in 
$$\frac{27!}{4!5!3!7!8!}$$
distinguishable ways.  
One pair of adjacent C's and two overlapping pairs of adjacent C's:  We have $27$ objects to arrange:  $4$ A's, $5$ B's, $1$ CC, $1$ CCC, $1$ C, $7$ D's, $8$ E's.  They can be arranged in 
$$\frac{27!}{4!5!1!1!1!7!8!}$$
distinguishable ways.
Three overlapping pairs of adjacent C's:  We have $27$ objects to arrange:  $4$ A's, $5$ B's, $1$ CCCC, $2$ C's, $7$ D's, $8$ E's.  They can be arranged in 
$$\frac{27!}{4!5!1!2!7!8!}$$
distinguishable ways.
Hence, 
$$|A_3| = \frac{27!}{4!5!3!7!8!} + \frac{27!}{4!5!1!1!1!7!8!} + \frac{27!}{4!5!1!2!7!8!}$$
Four pairs of adjacent C's
Two sets of two overlapping pairs of adjacent C's:  We have $26$ objects to arrange: $4$ A's, $5$ B's, $2$ CCC's, $7$ D's, $8$ E's.  They can be arranged in 
$$\frac{26!}{4!5!2!7!8!}$$
distinguishable ways.
One pair of adjacent C's and three overlapping pairs of adjacent C's:  We have $26$ objects to arrange:  $4$ A's, $5$ B's, $1$ CC, $1$ CCCC, $7$ D's, $8$ E's.  They can be arranged in 
$$\frac{26!}{4!5!1!1!7!8!}$$
distinguishable ways.
Four overlapping pairs of adjacent C's:  We have $26$ objects to arrange:  $4$ A's, $5$ B's, $1$ CCCCC, $1$ C, $7$ D's, $8$ E's.  They can be arranged in 
$$\frac{26!}{4!5!1!1!7!8!}$$
distinguishable ways.
Hence,
$$|A_4| = \frac{26!}{4!5!2!7!8!} + \frac{26!}{4!5!1!1!7!8!} + \frac{26!}{4!5!1!1!7!8!}$$
Five pairs of adjacent C's
We have $25$ objects to arrange:  $4$ A's, $5$ B's, $1$ CCCCCC, $7$ D's, $8$ E's.  They can be arranged in 
$$|A_5| = \frac{25!}{4!5!1!7!8!}$$
distinguishable ways.
Let $|B_k|$ denote the number of arrangements with at least $k$ pairs of adjacent C's.
Number of arrangements with at least one pair of adjacent C's
$$|B_1| = |A_1| - |A_2| + |A_3| - |A_4| + |A_5|$$
Number of arrangements with at least two pairs of adjacent C's
$$|B_2| = |A_2| - |A_3| + |A_4| - |A_5|$$
Number of arrangements with at least three pairs of adjacent C's
$$|B_3| = |A_3| - |A_4| + |A_5|$$
Number of arrangements with at least four pairs of adjacent C's
$$|B_4| = |A_4| - |A_5|$$
Number of arrangements with five pairs of adjacent C's
$$|B_5| = |A_5|$$
The number of arrangements with exactly one pair of adjacent C's is 
\begin{align*}
|B_1| & - |B_2| + |B_3| - |B_4| + |B_5|\\ 
& = (|A_1| - |A_2| + |A_3| - |A_4| + |A_5|) - (|A_2| - |A_3| + |A_4| - |A_5|) + (|A_3| - |A_4| + |A_5|) - (|A_4| - |A_5|) + |A_5|\\
& = |A_1| - 2|A_2| + 3|A_3| - 4|A_4| + 5|A_5|\\
& = 281625478590456000
\end{align*}