Art of Problem Solving

2019 Mock AMC 10B Problems/Problem 2: Difference between revisions

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AoPS: Bogus Solution
Simply choosing 3 of the boy scouts predetermines the other 3 boy scouts so we have <math>{6 \choose 3}=20</math>
<math>\boxed{\bold{D}}</math>
<math>\bold{20}</math>

Latest revision as of 07:52, 4 November 2025

Problem

Al, Bob, Clayton, Derek, Ethan, and Frank are six Boy Scouts that will be split up into two groups of three Boy Scouts for a boating trip. How many ways are there to split up the six boys if the two groups are indistinguishable?


$\textbf{(A)}\ 5 \qquad\textbf{(B)}\ 10  \qquad\textbf{(C)}\ 15 \qquad\textbf{(D)}\ 20 \qquad\textbf{(E)}\ 35$

Solution

There are ${6 \choose 3}=20$ ways to pick three boys. However, we have overcounted by double as if we choose the latter of three boys before its former counterpart, then we get the same case, so our answer is \( \frac{20}{2} = \) $\boxed{10}$.

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