Combinatorics: 6 indistinguishable balls and 6 different pair of balls I would appreciate some help with this

Within a box there are $6$ indistinguishable balls and $6$ different
  pairs of balls. 
In how many way can be chosen $6$ balls from the box?
In how many ways can be distributed among three different people?

My attempt for the first question:
First I separated the balls into two classes: Those that are unpaired (Class A, that happen to be indistinguishable) and those who are paired (Class B). Then I began considering the cases where we have:


*

*$0$ pairs from class B and $6$ from class A.

*$1$ pair from class B and $4$ balls from class A.

*$2$ pairs from B and $2$ from A.

*$3$ pairs from B and $0$ from A.


According to me: 
$$\binom{6}0\binom{6}6+\binom{6}1\binom{6}4+\binom{6}2\binom{6}2+\binom{6}3\binom{6}0 = 336$$
But the correct answer is $64$
My attempt for the 2nd question:
We can think on the people as three different boxes
This part can be separated into two. First we find the ways of distributing 6 indistinguishable balls in 3 different boxes. As fas as I understand this is reduced to a stars and bars problem. So we have
$$\binom{6+3-1}3$$
different ways.
For the second part we have $6$ different pairs of balls that have to be distribute in $3$ different boxes. First put the pairs in some arrange manner, labeled from $1$ to $6$. For the first there will be $3$ options. And the same is true for the rest of pairs. So we have $3^6$ ways of distributing this pairs among 3 different boxes. Then by the rule of sum we have
$$\binom{6+3-1}3 + 3^6 = 785$$
But the textbook says that is $20412$
 A: As I interpret the question, the box contains $18$ balls: $\{6a, 2b, 2c, 2d, 2e, 2f, 2g\}$. I would undertake this by inclusion/exclusion as follows:
First, choosing $6$ balls from $7$ categories has ${12 \choose 6} = 924$ options. However this includes a lot of cases where there are more than $2$ balls chosen from the latter six categories. So calculating how many ways to choose as above but with $3$ or more balls from category $b$, say, gives ${9 \choose 6} = 84$ options for that for each pair category. We also need the cases which would break two restrictions at once; the cases where we would select $3$ balls from two of the pair categories, which can only be done  ${6 \choose 2} = 15$ ways total.
Then combining this knowledge, we can strike the single-constraint breaking cases and add back in the double-counted double-constraint breaking cases:
$${12 \choose 6} - 6{9 \choose 6} + {6 \choose 2} = 924 - 6\cdot 84 +15 = 435$$
Of course this is not close to the book answer, or indeed your answer, but I leave it as an example of selection with constraints.
A: If you assume the balls as: aaaaaa, bb, cc, dd, ee, ff, gg (with the condition that any of bb, cc, dd, ee, ff, gg can only be selected as "take both or none").
As you said, there are four ways:


*

*select 6 a's. Result = $\binom{6}{0}$=1

*select 4 a's and rest others. Result = $\binom{6}{1}$=6

*select 2 a's and rest others. Result = $\binom{6}{2}$=15

*select 0 a's and rest others. Result = $\binom{6}{3}$=20
Total = 42.
For some reason, the expected answer 64 happens to be equal to $\binom{7}0+\binom{7}1+\binom{7}2+\binom{7}3$
