Five balls of different colours are to be placed in three boxes of different sizes. The number of ways in which we can place the balls in the boxes so that no box remains empty is
Let's denote the five balls by A,B,C,D,E and the three different boxes by X,Y,Z. We need to place these five balls into the three boxes such that no box remains empty. First, let's calculate the total number of ways to distribute 5 balls into 3 boxes without any restriction. Since each of the 5 balls can be placed in any of the 3 boxes, the number of unrestricted distributions is: 35=243
Now, we need to subtract the cases where one or more boxes are empty. We'll use the principle of inclusion-exclusion to account for these cases. If one box is empty, we need to distribute the balls into the remaining two boxes. There are: 25=32 ways to distribute the 5 balls into the 2 remaining boxes. Since there are 3 boxes and any one of them can be empty, we need to multiply by 3 : 3×32=96
Step 2: Add back the cases where two boxes are empty If two boxes are empty, all the balls must go into the remaining one box. There is only 1 way to do this, and since we can choose which single box is non-empty in 3 ways, we need to add back: By applying the principle of inclusion-exclusion, the number of distributions where no box is empty is: 243−96+3=150 Thus, the number of ways to place the five different-colored balls in three different boxes such that no box remains empty is: Option D: 150