As the comment of Karl indicates, neither Case 1 nor Case 2 is appropriate here. Although I could be mistaken, I strongly suspect that (as Karl's comment also indicates), in order to complete the problem, you have to re-work the first part of the problem. That is, you have to consider the mutually exclusive cases of how many distinct objects were taken from A and from B.
The (very tedious) work is shown below. I emphasize that I may well be overlooking a more elegant approach to the problem.
For selecting items from A :
There are $~\displaystyle (10)~$ ways of selecting the same item twice.
I will use the syntax [A:2] to represent this possibility.
There are $~\displaystyle \binom{10}{2} = 45~$ ways of selecting two distinct items.
I will use the syntax [A:1-1] to represent this possibility.
Note
$45 + 10 \neq 10^2 = 100,~$ and $~100~$ was reported by the OP (i.e. original poster). The discrepancy is explained by noting that when two distinct items are selected, the order of selection is irrelevant to the second part of the problem.
For selecting items from B :
There are $~\displaystyle \binom{6}{1} = 6~$ ways of selecting the same item, four times.
I will use the syntax [B:4] to represent this possibility.
There are $~\displaystyle (6 \times 5) = 30~$ ways of selecting one item three times, and then a second item one time.
I will use the syntax [B:3-1] to represent this possibility.
There are $~\displaystyle \binom{6}{2} = 15~$ ways of selecting one item twice, and then selecting a separate item twice.
I will use the syntax [B:2-2] to represent this possibility.
There are $~\displaystyle \binom{6}{2} \times 4 = 60~$ ways of selecting two distinct items, one time each, and then selecting a third item twice.
I will use the syntax [B:2-1-1] to represent this possibility.
There are $~\displaystyle \binom{6}{4} = 15~$ ways of selecting four distinct items.
I will use the syntax [B:1-1-1-1] to represent this possibility.
In effect, the above case work has partitioned the two positive integers $~4~$ and $~2.$
Now, in order to complete the problem, unfortunately, the $~2 \times 5 = 10~$ cases have to handled separately. To keep this answer as short as possible, I will leave the reader to review Multinomials on their own.
The basic idea is that if you have $~k~$ distinct items, and each item is selected $~a_i~$ times, where $~i \in \{1,2,\cdots,k\},~$ and if $~\displaystyle \sum_{i=1}^k a_i = n$:
then the number of distinct ways of arranging the items in a row is
$$\frac{n!}{(a_1)! \times (a_2)! \times \cdots \times (a_k)!} \tag1 $$
For ease of syntax, I will use the expression $~\displaystyle \binom{n}{a_1,a_2,\cdots,a_k}~$ to represent the expression in (1) above.
Then, the number of distinct distributions of the items in a row is given by summing each of the $~10~$ cases separately. This summation is detailed below:
[A:2] $~\times~$ [B:4]
$~:~~~~\displaystyle [10 \times 6] \times \binom{6}{2,4} ~$ ways.
[A:2] $~\times~$ [B:3-1]
$~:~~~~\displaystyle [10 \times 30] \times \binom{6}{2,3,1} ~$ ways.
[A:2] $~\times~$ [B:2-2]
$~:~~~~\displaystyle [10 \times 15] \times \binom{6}{2,2,2} ~$ ways.
[A:2] $~\times~$ [B:2-1-1]
$~:~~~~\displaystyle [10 \times 60] \times \binom{6}{2,2,1,1} ~$ ways.
[A:2] $~\times~$ [B:1-1-1-1]
$~:~~~~\displaystyle [10 \times 15] \times \binom{6}{2,1,1,1,1} ~$ ways.
[A:1-1] $~\times~$ [B:4]
$~:~~~~\displaystyle [45 \times 6] \times \binom{6}{1,1,4} ~$ ways.
[A:1-1] $~\times~$ [B:3-1]
$~:~~~~\displaystyle [45 \times 30] \times \binom{6}{1,1,3,1} ~$ ways.
[A:1-1] $~\times~$ [B:2-2]
$~:~~~~\displaystyle [45 \times 15] \times \binom{6}{1,1,2,2} ~$ ways.
[A:1-1] $~\times~$ [B:2-1-1]
$~:~~~~\displaystyle [45 \times 60] \times \binom{6}{1,1,2,1,1} ~$ ways.
[A:1-1] $~\times~$ [B:1-1-1-1]
$~:~~~~\displaystyle [45 \times 15] \times \binom{6}{1,1,1,1,1,1} ~$ ways.
Addendum
After reviewing Joriki's more elegant answer, I sanity-checked the two results. My final computation exactly agrees with his.
Since I don't believe in coincidences, I decided to stretch my intuition to understand $~\color{red}{why}~$ the computations agree.
For simplicity, consider the situation where two distinct items are selected from A and one item is selected four times from B.
My enumeration of this particular possibility is:
$$\binom{10}{2} \times \binom{6}{1} \times \binom{6}{1,1,4} = 45 \times 6 \times 30 = 8100.$$
Joriki's enumeration of this particular possibility is:
$$[10 \times 9] \times [6] \times \binom{6}{2} = 90 \times 6 \times 15 = 8100.$$
