How many ways are there of distributing 5 distinct objects into 4 boxes so that at least 2 boxes remain empty?

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Given:

6 distinct balls can be put in 5 distinct boxes.

Calculation:

The number of ways of in n distinct objects can be put into identical boxes, so that neither one of them remains empty.

Since both the boxes and the balls are different , we can choose any box , and every choice is different at any time.

The first ball can be placed in any of the 5 boxes . Similarly , the other balls can be placed in any of the 5 boxes.

The number of ways = 56 = 15625

∴ The number of ways is 15625.

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Your first part is correct. The number of the total ways is $4^5$.

Then you might consider the in-exclusion principle: $A_1$: $Box_1$ is empty; $A_2$: $Box_2$ is empty; $A_3$: $Box_3$ is empty; $A_4$: $Box_4$ is empty.

Use the same method that you got the number of total ways. For each of these cases, distributing $5$ distinguishable balls to $3$ distinguishable boxes, number of ways is $3^5$. $\binom{4}{1}$ shows it has $4$ cases.

$\vert A_1 \bigcap A_2\vert$ (Both Box 1 and Box 2 are empty) $= 2^5$. $\binom{4}{2}$ (from $A_1$ - $A_4$, choose $2$) shows how many cases in this situation.

$\vert A_1 \bigcap A_2 \bigcap A_3\vert$ (Boxes 1,2,3 are empty) $= 1^5$. $\binom{4}{3}$ cases.

$\vert A_1 \bigcap A_2 \bigcap A_3 \bigcap A_4 \vert$ (Boxes 1,2,3,4 are empty)= $0^5$. $\binom{4}{4}$ case.

Then apply inclusion-exclusion principle, $\vert A_1 \bigcup A_2 \bigcup A_3 \bigcup A_4 \vert$(at least one box is empty) $$=\sum_{k=1}^4 (-1)^{k+1}(4-k)^5 \binom{4}{k}$$

Finally, use the total number $4^5$ minus the number we got from the last step, then we get the number of ways that no box is empty.

To be honest, this method is much more complicated than other approaches. However, I guess you might want to generalise it to distribute $r$ distinguishable balls into $n$ distinguishable boxes(?) Then it definitely should be a good way to think about this question.