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

(Created page with "Simplify 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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Simplify 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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Simply choosing 3 of the boy scouts predetermines the other 3 boy scouts so we have <math>{6 \choose 3}=20</math>
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<math>\boxed{\bold{D}}</math>
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<math>\bold{20}</math>
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THE ABOVE SOLUTION IS WRONG!!!!
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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 16:32, 2 November 2024

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


THE ABOVE SOLUTION IS WRONG!!!!

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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