2019 Mock AMC 10B Problems/Problem 2

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

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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2019 Mock AMC 10B Problems/Problem 2