A different approach in distributing $8$ distinct balls into $6$ distinct boxes such that each box has at least $1$ ball Find the number of ways in distributing $8$ distinct balls into $6$ distinct boxes such that there is at least $1$ ball in each box.
We are well acquainted with the traditional Inclusion-Exclusion principle in solving this and it will come up as $191520$ ways.
I tried in this way:
When we distribute $8$ balls into $6$ boxes we have two possible cases
Case $1.$ There are exactly $2$ boxes with $2$ balls each
Case $2.$ There is exactly $1$ Box with $3$ balls in it.
For Case $1$, the distribution is done by choosing $2$ boxes from $6$ boxes and $2$ balls from $8$ balls and those two balls can be arranged in $2!$ ways. But I require another two balls from remaining $6$ balls to be placed in each of these two boxes which can be done in $\binom{6}{2} \times 2!$ ways. Finally remaining $4$ balls can be distributed in remaining $4$ boxes in $4!$ ways. So total number of ways for Case $1.$ is
$$\binom{6}{2} \times \binom{8}{2} \times 2! \times \binom{6}{2} \times 2! \times 4!=604800$$
For Case $2$, the number of ways is
$$\binom{6}{1} \times \binom{8}{3} \times 5!=40320$$
I am pretty sure the value obtained for Case $2.$ is correct. but I could not find where I went wrong in Case $1.$

On the case 1 you must choose first $2$ boxes over $6$ i.e. $C(6,2)$ and after for the first box you choose $(8\cdot 7)/2!$ and after $(6\cdot 5)/2!$ for the second box, and in third place you count the permutations of the others boxs that is $4!$, so the total is $\binom{6}{2}^2\cdot \binom{8}{2}\cdot4!=15^2\cdot 28\cdot 24=151200$.
And $151200+40320=191520$ as stated.
