how many option there are to put the letters AAAABBBBCCCC (4 A, 4 B, 4 C) in a word so that there are 2 A next to each other?
for example AAAABBBBCCCC counts as an option.
is there a way to think about it as a problem with cells and balls? without Inclusion-Exclusion Principle ?
Attempt: The number of arrangements of the letters AAAABBBBCCCC with no restrictions is $\binom{12}{4}\binom{8}{4}\binom{4}{4}$.
Best Answer
As you correctly calculated, there are $\binom{12}{4}\binom{8}{4}\binom{4}{4}$ ways to arrange the letters $AAAABBBBCCCC$ without restriction. From these, we must subtract those arrangements in which no two $A$s are adjacent. To do so, we first arrange the letters $BBBBCCCC$, which can be done in $\binom{8}{4}\binom{4}{4}$ ways. This creates nine spaces, seven between successive letters and two at the ends of the row where we can place the $A$s.
$$\square L \square L \square L \square L \square L \square L \square L \square L \square$$ where each $L$ represents one of the eight letters in the string $BBBBCCCC$. To ensure that no two $A$s are adjacent, we must select four of these nine spaces in which to place the $A$s. Hence, there are $\binom{8}{4}\binom{4}{4}\binom{9}{4}$ arrangements in which no two $A$s are adjacent. Thus, the number of admissible arrangements is $$\binom{12}{4}\binom{8}{4}\binom{4}{4} - \binom{8}{4}\binom{4}{4}\binom{9}{4}$$