Question

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 ways can we place the balls so that no box remains empty?

• A. $50$
• B. $100$
• C. $150$
• D. $200$

Answer 1

The correct option is C

$150$

Step 1: Find the number of possible arrangements if balls are arranged in order of $(2,2,1)$

We have been given, five balls of different colors to be placed in three boxes of different sizes. Each box can hold all five balls.

We need to find the number of ways we can place the balls so that no box remains empty

if balls are arranged in order $(2,2,1)$ then the number of possible arrangements would be,

$$\begin{gathered} =3\times {}^{5}C_2\times {}^{3}C_2\times {}^{1}C_1 \\ =3\times \frac{5!}{2!(5-2)!}\times \frac{3!}{2!(3-2)!}\times \frac{1!}{1!(1-1)!} \\ =3\times 10\times 3\times 1 \\ =90 \end{gathered}$$

Step 2: Find the number of possible arrangements if balls are arranged in order of $(1,1,3)$

If balls are arranged in order $(1,1,3)$ then the number of possible arrangements would be,

$$\begin{gathered} =3\times {}^{5}C_1\times {}^{4}C_1\times {}^{3}C_3 \\ =3\times \frac{5!}{1!(5-1)!}\times \frac{4!}{1!(4-1)!}\times \frac{3!}{3!(3-3)!} \\ =3\times 5\times 4\times 1 \\ =60 \end{gathered}$$

Step 3: Find the total number of arrangements.

From steps $\left(1\right)\mathrm{and}\left(2\right),$

the total number of arrangements would be,

$$\begin{gathered} =90+60 \\ =\mathbf{150} \end{gathered}$$

Therefore, option (C) is the correct answer.