Bunuel wrote:
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 place the balls so that no box remains empty?
A. 75
B. 90
C. 120
D. 150
E. 180
We have 5 different balls and 3 different boxes. There is at least one ball in each box, so the different combinations become very limited and can be worked out individually.
1) 3-1-1
This means that one of the box has 3 balls, while the other two have one eachWays to choose Balls : we can choose 3 balls out of 5 in 5C3 ways, and then choose 1 out of remaining 2 in 2C1 ways => 5C3*2C1*1C1=20 ways
Ways to choose boxes: we have 3 in one and two of them have similar number, that is 1. The permutations when 2 are similar are 3!/2!=3 ways.
Total ways = 20*3=60 ways
2) 2-2-1......The explanation is similar to the first case.
Balls in 5C2*3C2=30
boxes in 3!/2!=3 ways.
Total 30*3=90
Total 60+90=150
D