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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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We give an alternate approach that (barely) involves cases. It will give a number to check your answer against.
The box that gets $3$ balls can be chosen in $7$ ways, and for each way the balls it gets can be chosen in $\binom{7}{3}$ ways.
What about the rest? The remaining $4$ balls can, almost, be assigned in $6^4$ ways. Except that we do not want any of the remaining boxes to have $3$ balls.
How many bad distributions are there? The box that gets $3$ can be chosen in $6$ ways, and the balls it gets can be chosen in $\binom{4}{3}$ ways, with the remaining ball assigned in $5$ ways. Thus the total is $$7\binom{7}{3}\left(6^4 -(6)\binom{4}{3}(5)\right).$$
More or less equivalently, we can use Inclusion/Exclusion more explicitly. Call the boxes $1$ to $7$. For each $i$, there are $\binom{7}{3}6^4$ ways of assigning $3$ balls to Box $i$, with the rest distributed among the others.
If we add up, $i=1$ to $7$, we will have double-counted the situations where two boxes get $3$ balls. But we want to count these no times.
So from $7\cdot \binom{7}{3}\cdot 6^4$ we must subtract $2\cdot \binom{7}{2}\binom{7}{3}\binom{4}{3}\cdot 5$.