Question

A group consists of 4 girls and 7 boys. in how many ways a team of 5 members be selected if the team has (a) No girl (b) at least one girl and boy (c) atleast3 girls?

Answer 1

$$\begin{gathered} 4\mathrm{girls},7\mathrm{boys} \\ \mathrm{A}5\mathrm{member}\mathrm{team}\mathrm{to}\mathrm{be}\mathrm{formed}\mathrm{such}\mathrm{that} \\ \left(\mathrm{a}\right)\mathrm{No}\mathrm{girl} \\ \mathrm{All}\mathrm{members}\mathrm{to}\mathrm{be}\mathrm{boys}{=}^7{\mathrm{C}}_5=\frac{7!}{5!2!}=21\mathrm{ways} \\ \left(\mathrm{b}\right)\mathrm{at}\mathrm{least}1\mathrm{girl}\mathrm{and}1\mathrm{boy} \\ 1\mathrm{girl}4\mathrm{boy} \\ {\Rightarrow }^4{\mathrm{C}}_1{\times }^7{\mathrm{C}}_4=4\times 35=140 \\ 2\mathrm{girl}3\mathrm{boy} \\ {\Rightarrow }^4{\mathrm{C}}_2{\times }^7{\mathrm{C}}_3=6\times 35=210 \\ 3\mathrm{girl}2\mathrm{boy} \\ {\Rightarrow }^4{\mathrm{C}}_3{\times }^7{\mathrm{C}}_2=4\times 21=84 \\ 4\mathrm{girl}1\mathrm{boy} \\ {\Rightarrow }^4{\mathrm{C}}_4{\times }^7{\mathrm{C}}_1=1\times 7=7 \\ \mathrm{total}\mathrm{way}=140+210+84+7 \\ =441 \\ \left(\mathrm{c}\right)\mathrm{atleast}3\mathrm{girls} \\ 3\mathrm{girl}2\mathrm{boy} \\ {\Rightarrow }^4{\mathrm{C}}_3{\times }^7{\mathrm{C}}_2=4\times 21=84 \\ 4\mathrm{girl}1\mathrm{boy} \\ {\Rightarrow }^4{\mathrm{C}}_4{\times }^7{\mathrm{C}}_1=1\times 7=7 \\ \mathrm{total}\mathrm{ways}=84+7=91\mathrm{ways} \end{gathered}$$