GO Classes Scholarship 2023 | Test | Question: 6

Answer 1

Clearly, all the boxes are different. So, let's name them $\text{B1, B2, B3}.$ We need to put all balls in the boxes, so, $\text{B1}$ will have $3, \mathrm{~B} 2$ will have $4$ and $\text{B3}$ will have $5$ balls.
All red balls go in the same box: There are $3$ ways to do so.
Way 1: All red balls in Box $\text{B1} :$
Since all blue balls are distinct, so, we have $^9 \mathrm{C} _4$ ways to do so.
Way 2: All red balls in Box $\text{B2}:$
Since all blue balls are distinct, so, we have $9 \ast (^8 \mathrm{C}_3)$ ways to do so.
Way 3: All red balls in Box $\text{B3}:$
Since all blue balls are distinct, so, we have $^9 \mathrm{C}_2 \ast \; ^7 \mathrm{C}_3$ ways to do so.

So, the total is $1890.$

Detailed Video Solution:

https://youtu.be/tqjuxfutFHg?t=3475