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Given: 2 teachers, 7 boys, 4 girls

The question is: "If the committee (5 persons) must include at least one teacher and two boy students, in how may ways can the committee be selected?"

My question is if I can include more than two boys, or must there be exactly 2 boys.

Thank you.

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  • $\begingroup$ Exactly two boys $\endgroup$ – Upstart Mar 1 '16 at 16:26
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    $\begingroup$ Resolve both situations. I tend to see here "at least two boys". $\endgroup$ – Masacroso Mar 1 '16 at 16:39
  • $\begingroup$ inclusion and exclusion works here $\endgroup$ – ThunderWiring Mar 1 '16 at 16:57
  • $\begingroup$ The question is not clearly worded, but I tend to agree with Masacroso. $\endgroup$ – true blue anil Mar 1 '16 at 17:24
  • $\begingroup$ I agree with Masacroso that it's probably "at least two boys", but ask for clarification if possible. If not, work the problem for both cases, they're not that much different. $\endgroup$ – DylanSp Mar 1 '16 at 17:45
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Ask the author of the problem, whether "at least two boys" or "exactly two boys" is meant. MSE cannot help you in this dilemma.

In the following I'm treating the "at least two boys" interpretation.

We can take both teachers, $b\in\{2,3\}$ boys and $g=3-b$ girls in $${7\choose 2}{4\choose 1}+{7\choose 3}{4\choose0}=119$$ ways, and we can take $1$ teacher, $b\in\{2,3,4\}$ boys and $g=4-b$ girls in $$2{7\choose 2}{4\choose2}+2{7\choose3}{4\choose1}+2{7\choose4}{4\choose0}=602$$ ways, giving a total of $721$.

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Assuming that at least two boys are required: figure out how many different ways $1$ required teacher slot can be filled from $2$ different teachers, multiply that by the number of ways the $2$ slots for boys can be filled from $7$ different boys, then multiply that by the number of ways the remaining two slots can be filled from $(2 - 1)$ teachers plus $(7 - 2)$ boys plus $4$ girls. If exactly two boys are required, then disregard the boys when computing who can fill the last two spots.

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