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Difference between revisions of "2019 Mock AMC 10B Problems/Problem 2"
 
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Simply choosing 3 of the boy scouts predetermines the other 3 boy scouts so we have <math>{6 \choose 3}=15</math>
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==Problem==
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<cmath></cmath><math>\boxed{\bold{C}}</math>
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<math>\bold{25}</math>
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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?
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<math>\textbf{(A)}\ 5 \qquad\textbf{(B)}\ 10  \qquad\textbf{(C)}\ 15 \qquad\textbf{(D)}\ 20 \qquad\textbf{(E)}\ 35</math>
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==Solution==
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There are <math>{6 \choose 3}=20</math> 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 \( \frac{20}{2} = \) <math>\boxed{10}</math>.
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~Pinotation
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AoPS: Bogus Solution
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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>
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<math>\bold{20}</math>
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This solution is wrong because every time we choose 3 boys, we predetermine the other group. But if we instead end up choosing the other group, the first 3 boys are predetermined. In other words, if we have 6 boys, (A, B, C, D, E, F) and we choose A,B,C for one group we will have D,E,F for the other group. However, because the groups are indistinguishableable, if we choose D,E,F, for one group we will have A,B,C in the other group, which is just a duplicate case. Therefore, we will divide <math>{6 \choose 3}</math> by 2 and get <math>\frac{20}{2} = 10</math>, which is <math>\boxed{\bold{(B)} 10}</math> - lucaswujc

Latest revision as of 13:37, 21 September 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 \( \frac{20}{2} = \) $\boxed{10}$.

~Pinotation

AoPS: Bogus Solution

Simply choosing 3 of the boy scouts predetermines the other 3 boy scouts so we have ${6 \choose 3}=20$ $\boxed{\bold{D}}$ $\bold{20}$

This solution is wrong because every time we choose 3 boys, we predetermine the other group. But if we instead end up choosing the other group, the first 3 boys are predetermined. In other words, if we have 6 boys, (A, B, C, D, E, F) and we choose A,B,C for one group we will have D,E,F for the other group. However, because the groups are indistinguishableable, if we choose D,E,F, for one group we will have A,B,C in the other group, which is just a duplicate case. Therefore, we will divide ${6 \choose 3}$ by 2 and get $\frac{20}{2} = 10$, which is $\boxed{\bold{(B)} 10}$ - lucaswujc

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