# A different approach in distributing $8$ distinct balls into $6$ distinct boxes

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 $$1.$$

• Your description is somewhat unclear. You say "We are well acquainted with the traditional Inclusion Exclusion principle in solving this and it will come up as $191520$ ways". So how can the different value that you've obtained for case #$2$ can be correct??? Dec 4 '15 at 14:17

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.
Case 1 is wrong: you pick 2 boxes from 6, in $\binom{6}{2}$ ways. These boxes are different, say boxes $i < j$. First fill box $i$ with 2 balls out of 8, so $\binom{8}{2}$ ways. Then fill box $j$ with 2 out of 6, so $\binom{6}{2}$ ways. Then we are left $4!$ ways for the remaining 4 balls in the remaining 4 boxes.
It only matters that balls say 7 and 5 are in box $i$, there is no order within that box! So we get a number 4 times as small, and then it checks out with your other answer.