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Question

There are 5 different boxes and 7 different balls. All the 7 balls are to be distributed among 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 the same box?

A
1200
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B
15000
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C
3800
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D
none of these
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Solution

The correct option is B 15000
The required number of distribution = Total number of distribution - Number of distribution in which ball 2 and ball 4 are together.
Here the total number of distributions = 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 5C1×6C2×4! ways=1800.
5C1 Number of ways of selecting one box in which two balls are kept together.
6C2 Number of ways of selecting 2 balls out of 6 balls.
4! Number of ways of distribution of remaining 4 balls in remaining 4 boxes.
Hence, the required number of ways = 16800 - 1800 = 15000

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