Question

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

Answer 1

Answer:

540

Step-by-step explanation:

Hi,

Let B₁, B₂, B₃, B₄ and B₅ be the 5 distinct colors

Let X, Y and Z be 3 distinct boxes.

We need to place the colors such that no box remains empty,

Let us do the operation in 2 steps:

Step 1: Chose 3 different colors and place in each box

Let us choose 3 different colors , this can be done in

⁵C₃ ways = 10, and let us place these 3 colors in 3 distinct boxes,

since any of these color can go to any of the box, it could be

done in 3! ways.

Total number of ways of choosing 3 colors and placing them in

3 boxes can be done in 10*3! = 60 ways

Step 2: Place the remaining 2 colors in any of 3 boxes

Since these 2 colors can be placed in any one of the boxes and

since there are 3 distinct boxes, each color can be placed in 3

different ways.

So, placing both the colors can happen in 3*3 = 9 different ways

Total number of ways = total number of ways of placing selected

3 colors in 3 distinct boxes and placing the rest of the  colors in

any of the box which can happen in

60*9 = 540 ways

Hope, it helps !

Answer 2

Step-by-step explanation:

the correct answer is 150...