in Combinatory recategorized by
228 views
4 votes
4 votes
Consider three boxes and $12$ balls of the same size. We have $3$ indistinguishable red balls and $9$ distinguishable blue balls. The first box can fit at most three balls, the second box can fit at most four balls and the third box can fit at most five balls. We put the balls into the boxes such that all the red balls go into the same box. What is the total number of ways to put all the balls in the boxes?
in Combinatory recategorized by
228 views

1 Answer

6 votes
6 votes

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

edited by
Answer:

Related questions