QJEEAdv22Ma4c8b: JEE Advanced 2022 Mathematics - Class 11 JEEAdv22Ma4c8b Solution: Consider 4 boxes, where each box contains 3 red balls and 2…

Case 1 :

Among four bags from one bag 4 balls are taken. Number of ways of chosing one bag from 4 bags = ${}^4{C_1}$

From this bag number of ways of taking 4 balls are

(a) 3 Red and 1 Blue balls, which can be chosen in ${}^3{C_3} \times {}^2{C_1}$ ways.

(b) 2 Red and 2 Blue balls, which can be chosen in ${}^3{C_2} \times {}^2{C_2}$ ways.

$\therefore$ Total number of ways of choosing this 4 balls from this bag = ${}^4{C_1}$ $\left( {{}^3{C_3} \times {}^2{C_1} + {}^3{C_2} \times {}^2{C_2}} \right)$

Now, two balls are taken from remaining three bags.

From each bag two balls can be taken in ${{}^3{C_1} \times {}^2{C_1}}$ ways.

So, for three bags two balls can be taken in ${\left( {{}^3{C_1} \times {}^2{C_1}} \right)^3}$ ways.

$\therefore$ Total number of ways of chosing 10 balls from these four bags

= ${}^4{C_1}$ $\left( {{}^3{C_3} \times {}^2{C_1} + {}^3{C_2} \times {}^2{C_2}} \right)$ $\times$ ${\left( {{}^3{C_1} \times {}^2{C_1}} \right)^3}$

Case 2 :

Among four bags from two bags 3 balls are taken. Number of ways of chosing two bags from 4 bags = ${}^4{C_2}$

From each bag, number of ways of taking 3 balls are

(a) 2 Red and 1 Blue balls, which can be chosen in ${}^3{C_2} \times {}^2{C_1}$ ways.

(b) 1 Red and 2 Blue balls, which can be chosen in ${}^3{C_1} \times {}^2{C_2}$ ways.

So, for two bags three balls can be taken in ${\left( {{}^3{C_2} \times {}^2{C_1} + {}^3{C_1} \times {}^2{C_2}} \right)^2}$ ways.

$\therefore$ Total number of ways of choosing this 3 balls from this two bags = ${}^4{C_2}$ ${\left( {{}^3{C_2} \times {}^2{C_1} + {}^3{C_1} \times {}^2{C_2}} \right)^2}$

Now, two balls are taken from remaining two bags.

From each bag two balls can be taken in ${{}^3{C_1} \times {}^2{C_1}}$ ways.

So, for two bags two balls can be taken in ${\left( {{}^3{C_1} \times {}^2{C_1}} \right)^2}$ ways.

$\therefore$ Total number of ways of chosing 10 balls from these four bags

= ${}^4{C_2}$ ${\left( {{}^3{C_2} \times {}^2{C_1} + {}^3{C_1} \times {}^2{C_2}} \right)^2}$ $\times$ ${\left( {{}^3{C_1} \times {}^2{C_1}} \right)^2}$

From Case 1 and Case 2, total number of ways of chosing 10 balls from these 4 boxes so that from each box at least one red ball and one blue ball are chosen

= ${}^4{C_1}$ $\left( {{}^3{C_3} \times {}^2{C_1} + {}^3{C_2} \times {}^2{C_2}} \right)$ $\times$ ${\left( {{}^3{C_1} \times {}^2{C_1}} \right)^3}$

+ ${}^4{C_2}$ ${\left( {{}^3{C_2} \times {}^2{C_1} + {}^3{C_1} \times {}^2{C_2}} \right)^2}$ $\times$ ${\left( {{}^3{C_1} \times {}^2{C_1}} \right)^2}$

$\begin{aligned}& =4(5)(6)^3+6(3 \times 2+3)^2(6)^2 \\\\& =4320+17496 \\\\& =21816\end{aligned}$