How many ways 6 different balls can be put in 6 different boxes such that exactly one box is empty a box can contain as many balls and every ball has to be put in a box?

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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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Assume that $a$ balls go to first box, $b$ to second and so on and $f$ to the sixth box.

Now, we have the equation $a+b+c+d+e+f=20$. But your condition says exactly $1$ must be $0$. So we put one variable to $0$ in the above equation and ensure that all the other variables are $\geq 1$.

So let's say $f=0$, so we have $a+b+c+d+e=20$ where $ a,b,c,d,e \ge 1 $.

The solution to this can be easily found by forming a generating function $(x+x^2+x^3+\cdots)^5$ in which we have to find the coeffcient of $x^{20}$. That can be easily found now.

Or, if one doesn't wants to make a generating function, we use stars and bars method here. Let $a'=a+1,b'=b+1,\cdots,e'=e+1$ so that $a',b',\cdots,e'$ are now $\geq 0$ and after plugging these variables into our equation, we get our new equation as $a'+b'+c'+d'+e'=15$, whose solution by stars and bars is given by $\displaystyle \binom{15+4}{4}=\binom{19}{4}$.

But since we assumed $f=0$, we could have any of $a,b,c,d,e$ to be $0$, so we multiply this number by $6$, thus final answer turns out to be $\displaystyle 6\binom{19}{4}$.

Note that $a'$ is not to be confused with a derivative (see comments), it's just another variable!