out of 8 students (4 boys and 4 girls), a taem of 4 is to be formed with atleast 2 boys and 2 girls in each team . in how many ways can this be done  

1. ​if riha is always to be included
2. if sanya is never to be taken

Dear Student,
Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{Language} \mathrm{of} \mathrm{question} \mathrm{is} \mathrm{poorly} \mathrm{constructed}. \\ \mathrm{Now} \mathrm{I}\hspace{0.17em}\mathrm{assume} \mathrm{that} \mathrm{there} \mathrm{must} \mathrm{be} \mathrm{atleast} 2 \mathrm{boys} \mathrm{and} \mathrm{no} \mathrm{such} \mathrm{restriction} \\ \mathrm{is} \mathrm{there} \mathrm{for} \mathrm{girls}. \\ 4 \mathrm{boys} \mathrm{and} 4 \mathrm{girls} \mathrm{are} \mathrm{there} \mathrm{and} \mathrm{team} \mathrm{of} 4 \mathrm{is} \mathrm{to} \mathrm{be} \mathrm{formed}. \\ 1. \mathrm{When} \mathrm{Riha} \mathrm{is} \mathrm{always} \mathrm{included} \\ \mathrm{As} \mathrm{Riha} \mathrm{is} \mathrm{included}, \mathrm{hence} \mathrm{we} \mathrm{have} 4 \mathrm{boys} \mathrm{and} 3 \mathrm{girls} \mathrm{and} \\ \mathrm{we} \mathrm{need} 3 \mathrm{more} \mathrm{members} \mathrm{such} \mathrm{that} \mathrm{atleast} \mathrm{two} \mathrm{of} \mathrm{them} \mathrm{are} \mathrm{boys}. \\ \mathrm{Number} \mathrm{of} \mathrm{ways}=\mathrm{Two} \mathrm{boys} \mathrm{and} \mathrm{One} \mathrm{girl} \mathrm{OR} \mathrm{Three} \mathrm{boys} \\ {=}^4{\mathrm{C}}_2{\times }^3{\mathrm{C}}_1{+}^4{\mathrm{C}}_3 \\ =\frac{4.3}{2.1}\times 3+\frac{4.3.2}{3.2.1} \\ =18+4 \\ =22 \\ 2. \mathrm{When} \mathrm{Sanya} \mathrm{is} \mathrm{never} \mathrm{selected}. \\ \mathrm{So} \mathrm{we} \mathrm{have} 4 \mathrm{boys} \mathrm{and} 3 \mathrm{girls}. \\ \mathrm{Number} \mathrm{of} \mathrm{ways}=\mathrm{Two} \mathrm{boys} \mathrm{and} \mathrm{Two} \mathrm{girls} \mathrm{OR} \mathrm{Three} \mathrm{boys} \mathrm{and} \mathrm{One} \mathrm{Girl} \mathrm{OR} \mathrm{Four} \mathrm{boys} \\ {=}^4{\mathrm{C}}_2{\times }^3{\mathrm{C}}_2{+}^4{\mathrm{C}}_3{\times }^3{\mathrm{C}}_1{+}^4{\mathrm{C}}_4 \\ =\frac{4.3}{2.1}\times \frac{3.2}{2.1}+\frac{4.3.2}{3.2.1}\times 3+1 \\ =6\times 3+4\times 3+1 \\ =31 \end{gathered}$$

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