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Question Number 102800 by bramlex last updated on 11/Jul/20

$$\mathrm{There}\mathrm{are}14\mathrm{boys}\mathrm{and}10\mathrm{girls}\mathrm{in}$$$$\mathrm{a}\mathrm{classroom}.\mathrm{The}\mathrm{teacher}\mathrm{wants}$$$$\mathrm{to}\mathrm{form}\mathrm{a}\mathrm{team}\mathrm{of}5\mathrm{students}.$$$$\mathrm{The}\mathrm{team}\mathrm{must}\mathrm{have}\mathrm{a}\mathrm{least}\mathrm{two}$$$$\mathrm{boys}\mathrm{and}\mathrm{two}\mathrm{girls}.\mathrm{Find}\mathrm{the}$$$$\mathrm{number}\mathrm{of}\mathrm{ways}\mathrm{the}\mathrm{team}\mathrm{can}\mathrm{be}$$$$\mathrm{chosen}.$$

Answered by bemath last updated on 11/Jul/20

$$case\left(1\right)3boys\mathrm{\&}2girls\Rightarrow C_3^{14}\times C_2^{10}$$$$=\frac{14.13.12}{3.2.1}\times \frac{10.9}{2.1}=13680$$$$\left(2\right)2boys\mathrm{\&}3girls\Rightarrow C_2^{14}\times C_3^{10}=\frac{14.13}{2.1}\times \frac{10.9.8}{3.2.1}$$$$=10920$$$$totally=13680+10920=24600$$$$$$

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