Question

There are 5 different boxes and 7 different balls. All the 7 balls are to be distributed in the 5 boxes placed in a row so that any box can receive any number of balls.

In how many ways can these balls be distributed such that no box is empty and ball 2 and ball 4 cannot be put in same box?

• A. 1200
• B. 15000
• C. 3800
• D. none of these

Answer 1

The correct option is B 15000

The required number of distributions = Total number of distributions - The number of distributions in which ball 2 and ball 4 are together.
Here the total number of distribution (as calculated in question number 68) = 16800.
Now, consider ball 2 and ball 4 are stuck together and this arrangement is assumed to be a single ball.
Thus we have 6 balls to be distributed into 5 boxes.Which can be done as 1, 1, 1, 1, 2 in ${}^5{C}_1{\times }^6{C}_2\times 4!ways=1800.$
${}^5{C}_1\to Numberofwaysofselectingoneboxinwhichtwoballsarekepttogether.$
${}^6{C}_2\to$ Number of ways of selecting 2 balls out of 6 balls.
$4!\to$ Number of ways of distribution of remaining 4 balls in remaining 4 boxes.
Hence, the required number of ways = 16800 - 1800 = 15000.