Difference between revisions of "2019 Mock AMC 10B Problems/Problem 2"
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| − | + | ==Problem== | |
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| − | <math>\ | + | 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 | ||
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?
Solution
There are 
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 \( \frac{20}{2} = \) 
.
~Pinotation
