Counting ways to arrange $5$ different balls into $3$ different boxes so that no box remains empty. I get $150$; official answer is $720$.

combinatorics

In how many ways can $5$ different balls be arranged into $3$ different boxes so that no box remains empty?

My Approach: Acc to me, since there are $2$ ways to arrange the balls ie,
$(1,1,3)$ and $(1,2,2)$ the total number of methods would be
$$ \frac{C^{5}_1\cdot C^4_1\cdot C^3_3 \cdot 3!}{2!} +\frac{C^{5}_1\cdot C^4_2\cdot C^2_2 \cdot 3!}{2!} = 150$$
By arranging the balls and then multiply by $3!$ as amount of ways $3$ different boxes can be arranged and dividing by $2!$ because one value is same.

Correct Answer: However, the correct answer is given as $$\frac{5! \cdot 3!}{2!}+ \frac{5! \cdot 3!}{2!}$$

What is wrong with my approach and why does the answer vary?

Edit: After reading all the answers, I also think that the question is not stated clearly and the order within the box must be important. It was taken from Skills In Mathematics algebra by Dr. SK Goyal (Chapter Permutation and Combination) if someone wants to look further upon the question.

Best Answer

If the order within the boxes matters, then the solution can be obtained quite easily:

Imagine that the balls are lined up like that: $\circ \: \circ \: \circ \: \circ \: \circ$

  • number of arrangements of $\color{blue}{}5$ balls: $\color{blue}{5!}$
  • putting in boxes with at least one ball per box corresponds to putting $\color{blue}{2}$ separators into the $\color{blue}{4}$ gaps between the lined up balls: $\color{blue}{\binom{4}{2}}$

All together:

$$\color{blue}{5!}\cdot \color{blue}{\binom{4}{2}} = \boxed{720}$$

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