Question

A group consists of 4 girls and 7 boys. In how many ways can a team of 5
members be selected if the team has
(i) no girls
(ii) at least one boy and one girl
(iii) at least three girls.

Answer 1

Dear Student
$$\begin{gathered} 4Gand7B \\ \left(i\right)nogirls \\ Numberofway={}^{7}C_5={}^{7}C_2=\frac{7\times 6}{2\times 1}=7\times 3=21 \\ \left(ii\right)atleastoneboyandonegirl \\ Numberofway=(1B\&4G)or(2B\&3G)or(3B\&2G)or(4B\&1G) \\ ={}^{7}C_1\times {}^{4}C_4+{}^{7}C_2\times {}^{4}C_3+{}^{7}C_3\times {}^{4}C_2+{}^{7}C_4\times {}^{4}C_1 \\ =7\times 1+\left(\frac{7\times 6}{2\times 1}\right)\times 4+\left(\frac{7\times 6\times 5}{3\times 2\times 1}\right)\left(\frac{4\times 3}{2\times 1}\right)+\left(\frac{7\times 6\times 5}{3\times 2\times 1}\right)\times 4 \\ =7+21\times 4+35\times 6+35\times 4=7+84+210+140 \\ =441 \\ \left(iii\right)atleastthreegirls \\ Numberofway=(2B\&3G)or(1B\&4G) \\ ={}^{7}C_2\times {}^{4}C_3+{}^{7}C_1\times {}^{4}C_4=\left(\frac{7\times 6}{2\times 1}\right)\times 4+7\times 1 \\ =21\times 4+7=84+7 \\ =91 \end{gathered}$$
Regards