In a high school, a committee has to be formed from a group of 6 boys $M_{1}, M_{2}, M_{3}, M_{4}, M

Question

In a high school, a committee has to be formed from a group of 6 boys $M_{1}, M_{2}, M_{3}, M_{4}, M_{5}, M_{6}$ and 5 girls $G_{1}$, $G_{2}, G_{3}, G_{4}, G_{5}$. (i) Let $\alpha_{1}$ be the total number of ways in which the committee can be formed such that the committee has 5 members, having exactly 3 boys and 2 girls. (ii) Let $\alpha_{2}$ be the total number of ways in which the committee can be formed such that the committee has at least 2 members, and having an equal number of boys and girls. (iii) Let $\alpha_{3}$ be the total number of ways in which the committee can be formed such that the committee has 5 members, at least 2 of them being girls. (iv) Let $\alpha_{4}$ be the total number of ways in which the committee can be formed such that the committee has 4 members, having at least 2 girls and such that both $M_{1}$ and $G_{1}$ are NOT in the committee together.   The correct option is :

Professor PC Thomas Classes · Answer

(c) : $\alpha_{1}={ }^{6} C_{3} \cdot{ }^{5} C_{2}=200$\ $\alpha_{2}={ }^{6} C_{1} \cdot{ }^{5} C_{1}+{ }^{6} C_{2} \cdot{ }^{5} C_{2}+{ }^{6} C_{3} \cdot{ }^{5} C_{3}+{ }^{6} C_{4} \cdot{ }^{5} C_{4}$\ $+{ }^{6} C_{5} \cdot{ }^{5} C_{5}=461$\ $\alpha_{3}={ }^{11} C_{5}-{ }^{6} C_{5}-{ }^{6} C_{4} \cdot{ }^{5} C_{1}={ }^{11} C_{5}-6-75$\ $={ }^{11} C_{5}-81=381$\[0pt] [ $G_{1}$ include ${ }^{4} C_{1} \cdot{ }^{5} C_{2}+{ }^{4} C_{2}{ }^{5} C_{1}+{ }^{4} C_{3}=74$\ $M_{1}$ include ${ }^{4} C_{2} \cdot{ }^{5} C_{1}+{ }^{4} C_{3}=34$\ $G_{1}$ and $M_{1}$, both not there\ ${ }^{4} C_{4}+{ }^{4} C_{3} \cdot{ }^{5} C_{1}+{ }^{4} C_{2} \cdot{ }^{5} C_{2}=81$\ $\therefore$ The total number of ways $\left(\alpha_{4}\right)$\ $=74+34+81=189$\