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 atleast $2$ 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, atleast $2$ 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, atleast $2$ girls and such that both $M_1$ and $G_1$ are NOT in the committee together.
$\begin{array}{cccc}& LIST-I& & LIST-II\\ P.& \text{The value of}{\alpha }_1\text{is}& 1.& 136\\ Q.& \text{The value of}{\alpha }_2\text{is}& 2.& 189\\ R.& \text{The value of}{\alpha }_3\text{is}& 3.& 192\\ P.& \text{The value of}{\alpha }_4\text{is}& 4.& 200\\ & & 5.& 381\\ & & 6.& 461\end{array}$
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

The correct option is C $P\to 4;Q\to 6;R\to 5;S\to 2$

$\left(i\right)$ committee has $5$ members, having exactly $3$ boys and $2$ girls$\left({\alpha }_1\right)$
$\therefore {\alpha }_1={}^6{C}_3\times {}^5{C}_2=200$
$\left(ii\right)$ Atleast $2$ members and equal number of boys and girls $\left({\alpha }_2\right)$
$1G1B\to {}^6{C}_1.{}^5{C}_1=302G2B\to {}^6{C}_2.{}^5{C}_2=1503G3B\to {}^6{C}_3.{}^5{C}_3=2004G4B\to {}^6{C}_4.{}^5{C}_4=755B5G\to {}^6{C}_5.{}^5{C}_5=6$
$\therefore {\alpha }_2=461$
$\left(iii\right)$committee has $5$ members, atleast $2$ of them being girls$\left({\alpha }_3\right)$
$2G3B\to {}^6{C}_3.{}^5{C}_2=2003G2B\to {}^6{C}_2.{}^5{C}_3=1504G1B\to {}^6{C}_1.{}^5{C}_4=305G0B\to {}^6{C}_0.{}^5{C}_5=1$
$\therefore {\alpha }_3=381$
$\left(iv\right)$committee has $4$ members, atleast $2$ girls both $M_1$ and $G_1$ are not in the committee together$\left({\alpha }_4\right)$$=$committee has $4$ members with atleast $2$ girls$-$ committee has $4$ members, atleast $2$ girls with both $M_1$ and $G_1$
$=({}^5{C}_2\cdot {}^6{C}_2+{}^6{C}_1\cdot {}^5{C}_3+{}^6{C}_0\cdot {}^5{C}_4)-({}^4{C}_1\cdot {}^5{C}_1+{}^4{C}_2)=189$