Five balls of different colors are to be placed in three different box

When you choose with \(C^2_4\) balls for the second box, the remaining 2 automatically go to the third box, so non need to multiply by 2 here.

Solution:

First, let’s determine the number of ways 3 positive integers can add up to 5:

1 + 1 + 3 = 5 and 1 + 2 + 2 = 5

Notice the number of addends is the number of boxes and the sum is the number of balls.

We see that there are two options. Since the boxes are different, for each option above, there are 3!/2! = 3 ways to arrange the addends.

Furthermore, for the first option, we have 5 choices to put a ball in the first box and 4 choices to put another ball in the second box (then the last 3 balls must go to the last box). Therefore, for the first option, we have 5 x 4 = 20 ways to place the 5 balls in the 3 boxes (if we ignore the ways in which we can arrange the boxes (or addends) for the moment).

For the second option, we have 5 choices to put a ball in the first box and 4C2 = 6 choices to put 2 balls in the second box (then the last 2 balls must go to the last box). Therefore, for the second option, we have 5 x 6 = 30 ways to place the 5 balls in the 3 boxes (again, let’s ignore the ways in which we can arrange the boxes (or addends) for the moment).

Now, if we also take the arrangement of the boxes into consideration, then the total number of ways we can place 5 different colored balls in 3 different boxes is:

20 x 3 + 30 x 3 = 60 + 90 = 150

Answer: D