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Mathematics 7 Online
OpenStudy (anonymous):

There are 5 boys and 5 girls on a co-ed basketball team. A team plays 5 players on the court at one time. How many possible ways are there to field a team of 3 boys and 2 girls? Position doesn’t matter.

OpenStudy (bradely):

5c3*5c2/10c5 Source: http://www.mathskey.com/question2answer/

OpenStudy (kropot72):

The number of combinations of the 5 boys taken 3 at a time is: \[\large C(5,\ 3)=\frac{5\times4}{2}\] The number of combinations of the 5 girls taken 2 at a time is: \[C(5,\ 2)=\frac{5\times4}{2}\] Each of the combinations of boys can be combined with each of the combinations of girls, therefore the total possible ways to field a team is given by: \[\large \frac{5\times4\times5\times4}{2\times2}=you\ can\ calculate\]

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