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!!