Five balls of different colors are to be placed in three boxes of different sizes. Each box can hold all five balls. In how many different ways can we we place the balls so that no box remains empty?

A. 150 B. 10 C. 60 D. 300 E. 375

I solved the question the following way.

3 balls can be selected in 5C3 ways. These three balls can be placed in three different boxes in 3! ways.
Hence total number of ways = 3! * 5C3

Each of remaining 2 balls can be placed in three boxes in 3 ways.

Hence final answer = 3! * 5C3 * 3 * 3 = 90.

But OA is A. i.e 150.

Could someone help? What am i missing here?

Many thanks in advance!!

U have to account for one more possibility that is the boxes are packed with the balls in the pattern 2,2,1 which will be 60 ways.

Therefore the correct answer is indeed 150(90+60) and the OA is correct!!!

please post the source of this question. if you do not do so, we will be forced to delete the question, to protect ourselves from the possibility of having posted questions from banned sources.

thank you.