Eight different dolls are to be packed with a condition Eight different dolls are to be packed  in eight different boxes. If two of the boxes are too small to hold five of the dolls, in how many ways can the dolls be packed? I tried to solve it, but got confused.
 A: 
Eight different dolls are to be packed in eight different boxes. If two of the boxes are too small to hold five of the dolls, in how many ways can the dolls be packed?

I'll assume this means that every doll goes into a unique box.
We can find this number using inclusion-exclusion:


*

*We include all $8!$ permutations of the dolls into boxes.

*We exclude the [???] permutations where small-box-1 gets a too-big doll.  We also exclude the [???] permutations where small-box-2 gets a too-big doll.

*We then re-include the [???] permutations where small-box-1 and small-box-2 get a too-big doll.
This gives [???] permutations.  We can verify this using the GAP code
Number(PermutationsList([1..8]),i->not (i[1] in [1..5] or i[2] in [1..5]));

by labeling the too-small boxes 1 and 2, and the too-big dolls 1 through 5.
Spoiler alert:

 We exclude $5 \cdot 7!$ twice (choosing one of the too-small boxes, and putting in it one of the 5 too-big dolls; the remainder are filled arbitrarily), and we re-include $5 \cdot 4 \cdot 6!$ (choosing both of the two-small boxes, and putting in them two of the too-big dolls; the remainder are filled arbitrarily).  This gives $8!-2 \cdot 5 \cdot 7! + 5 \cdot 4 \cdot 6!=4320$.

A: Consider $A_1, A_2,..,A_8 $ are the boxes where $b_1,b_2,..,b_8$ are the balls. Moreover $A_1$ and $A_2$ are too small to hold balls  $b_1,b_2,..,b_5$. Then, these $5$ balls goes to other $6$ boxes i.e. $A_3, A_4,..,A_8 $ in $6*5*4*3*2 = 720$ ways. Now there are $3$ balls left and $3$ boxes left. These balls can be arranged in $3! = 6$ ways. Hence total number of ways to arrange the balls is $720*6= 4320$
