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>
 
Simply choosing 3 of the boy scouts predetermines the other 3 boy scouts so we have <math>{6 \choose 3}=15</math>
<cmath>\boxed{\bold{C}}</cmath>
+
<cmath></cmath><math>\boxed{\bold{C}}</math>
<cmath>\bold{25}</cmath>
+
<math>\bold{15}</math>

Revision as of 11:43, 20 January 2020

Simply choosing 3 of the boy scouts predetermines the other 3 boy scouts so we have ${6 \choose 3}=15$ \[\]$\boxed{\bold{C}}$ $\bold{15}$