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.

\text{List-I} & \text{List-II} \\\ \text{P}\text{.The value of}~{{\alpha }_{1}} & \text{1}.\text{136} \\\ \text{Q}\text{.The value of}~{{\alpha }_{2}} & \text{2}.\text{189} \\\ \text{R}\text{.The value of}~{{\alpha }_{3}} & \text{3}.\text{192} \\\ \text{S}\text{.The value of}~{{\alpha }_{4}} & \text{4}.\text{2}00 \\\ {} & \text{5}.\text{381} \\\ {} & \text{6}.\text{461} \\\ \end{matrix}$$ The correct option is A.$P\to 4,Q\to 6,R\to 2,S\to 1$$$$$ B. $P\to 1,Q\to 4,R\to 2,S\to 3$$$$$ C. $P\to 4,Q\to 6,R\to 5,S\to 2$$$$$ D. $P\to 4,Q\to 2,R\to 3,S\to 1$$$$$

Answer 1

Use the formula for combination and fundamental principle of counting to find out the values of ${{\alpha }_{1}},{{\alpha }_{2}},{{\alpha }_{3}},{{\alpha }_{4}},{{\alpha }_{5}}$ using selection from the groups of girl and boys. Then you can match the options with list-I and list-II

Complete step-by-step answer:
We know that the selection of $r$ entities from $n$ unique entities is given by $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$. We expand in both denominator and numerator as

& ^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!} \\\ & =\dfrac{n\left( n-1 \right)...\left( n-r+1 \right)\left( n-r \right)...1}{r!\left( n-r \right)!} \\\ & =\dfrac{n\left( n-1 \right)...\left( n-r+1 \right)\left( n-r \right)!}{r!\left( n-r \right)!} \\\ & =\dfrac{n\left( n-1 \right)...\left( n-r+1 \right)}{r!} \\\ \end{aligned}$$ We also know by fundamental principle of counting selection of $r$ entities from $n$ and then $q$ entities from $m$ is given by $^{m}{{C}_{q}}^{n}{{C}_{r}}$. So the number of groups that can be formed when we form a group of ,$$$$ (i)5 members having 3 girls and 2 boys can be explained with the choice of 3 girls from the given 6 girls ( here $n=6,r=3$) simultaneously with the choice of 2 boys from the given 5 boys ( here $m=5,q=2$) . Then using the formula of $^{n}{{C}_{r}}$, we get $^6{C_3}^5{C_2} = \dfrac{{6.5.4}}{{3!}}\dfrac{{5.4}}{{2!}} = \dfrac{{6.5.4.5.4}}{{1.2.31.2}} = 200$ $$$$ (ii) at least 2 members, and having an equal number of boys and girls. We can find out possible combinations of at least one boy and one girl and continue to increase by one for both boys and girls $^{6}{{C}_{1}}^{5}{{C}_{1}}{{+}^{6}}{{C}_{2}}^{5}{{C}_{2}}{{+}^{6}}{{C}_{3}}^{5}{{C}_{3}}{{+}^{6}}{{C}_{4}}^{5}{{C}_{4}}{{+}^{6}}{{C}_{5}}^{5}{{C}_{5}}=461$ . So $Q\to 6$ $$$$ (iii) 5 members, at least 2 of them being girls is $^{6}{{C}_{3}}^{5}{{C}_{2}}{{+}^{6}}{{C}_{2}}^{5}{{C}_{3}}{{+}^{6}}{{C}_{1}}^{5}{{C}_{4}}{{+}^{6}}{{C}_{1}}^{5}{{C}_{5}}=381$. Here if we take girls rest will be boys and then continue to increase the number of girls by one. So $R\to 5$. $$$$ (iv) 4 members, having at least 2 girls and such that both ${{M}_{1}}$ and ${{G}_{1}}$ are NOT in the committee together is the sum of differences between all possible selection and selections of ${{M}_{1}}$ and ${{G}_{1}}$. That is $^{6}{{C}_{2}}^{5}{{C}_{2}}{{-}^{5}}{{C}_{1}}^{4}{{C}_{1}}{{+}^{6}}{{C}_{1}}^{5}{{C}_{3}}{{-}^{1}}{{C}_{1}}^{4}{{C}_{2}}{{+}^{5}}{{C}_{4}}=189$. So $S\to 2$.$$$$ **So, the correct answer is “Option C”.** **Note:** The proper selection is important while solving simultaneous combinatorial problems of this type especially in the case(iv) where we have to deal with special case exclusion because we have to take the difference from the total.