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Difference between revisions of "2021 April MIMC 10 Problems/Problem 4"

(Created page with "Stiskwey wrote all the possible permutations of the letters <math>AABBCCCD</math> (<math>AABBCCCD</math> is different from <math>AABBCCDC</math>). How many such permutations a...")
 
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==Solution==
 
==Solution==
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To be Released on April 26th.
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Use the theorem of over-counting (When arrange <math>n</math> distinguishable items and <math>m</math> indistinguishable items, the total number of ways to arrange them is <math>\frac{(n+m)!}{m!}</math>.) Therefore, the number of permutations of AABBCCCD is <math>\frac{8!}{2!2!2!}=\fbox{\textbf{(D)} 1680}</math>.

Latest revision as of 13:30, 26 April 2021

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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