2021 April MIMC 10 Problems/Problem 4

Stiskwey wrote all the possible permutations of the letters $AABBCCCD$ ($AABBCCCD$ is different from $AABBCCDC$). How many such permutations are there?

$\textbf{(A)} ~420 \qquad\textbf{(B)} ~630 \qquad\textbf{(C)} ~840 \qquad\textbf{(D)} ~1680 \qquad\textbf{(E)} ~5040$

Solution

Use the theorem of over-counting (When arrange $n$ distinguishable items and $m$ indistinguishable items, the total number of ways to arrange them is $\frac{(n+m)!}{m!}$.) Therefore, the number of permutations of AABBCCCD is $\frac{8!}{2!2!2!}=\fbox{\textbf{(D)} 1680}$.

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