I have 6 items. I need to select two but there are two sets of two items that I can only select one of. I have 6 items. I need to select two of these 6 items but there are two sets of two items in this set that I can only select one of. I do not necessarily need to select an item from either of these groups but if I do, I can only pick one from each. Can anyone assist me with figuring out the number of combinations of selections from the data set can I make? Any help would be greatly appreciated :)
For context, I have first counted the selections of any 2 from all 6 then I have tried to subtract the selections of what is incompatible from the two sets. I have performed that twice to account for the two sets but that doesn't feel right to me.
$$\binom{6}{2} - \binom{2}{1} - \binom{2}{1} = 11$$
 A: First you can take the situation where you can take all of the combinations out of the 6 items. Then you have a combination of 2 out of 6 which equals:
$\frac{6!}{(6-2)!*2!} = \frac{720}{48} = 15$
But there are two combinations that you can't choose: the combinations where you take two items of a set where you can't take two of.
If you take those in account, you find that you have 15-2 = 13 possible combinations.
I hope I did not overlook anything here.
A: There is only one way to take both people from the same group.
Method 1:  We use complementary counting.
There are $\binom{6}{2}$ ways to choose two of the six people in the group.  From these, we must subtract the number of ways we could take both people from one of the groups from which we are prohibited from taking both people.  There are two ways to choose the group from which we take both people and one way to choose both people in that group.  Hence, the number of admissible selections is $$\binom{6}{2} - \binom{2}{1}\binom{2}{2}$$
Method 2: We count directly.
To make this concrete, say the two groups from which only one person can be drawn be $A$ and $B$.  Let the remaining two people be in group $C$.
There are two possibilities:

*

*We choose one person each from two of the three groups.

*We take both people in group $C$.

We choose one person each from two of the three groups:  There are $\binom{3}{2}$ ways to select the two groups from which we take one person each and two ways to pick a person from each of those groups.  Hence, there are
$$\binom{3}{2}\binom{2}{1}\binom{2}{1}$$
such selections.
We choose both people from group $C$:  There is one way to pick both people from group $C$.
Total:  Since the above case are mutually exclusive and exhaustive, the number of admissible selections is
$$\binom{3}{2}\binom{2}{1}\binom{2}{1} + \binom{2}{2}$$
Alternatively, we could express the answer as
$$\binom{2}{1}\binom{2}{1}\binom{2}{0} + \binom{2}{1}\binom{2}{0}\binom{2}{1} + \binom{2}{0}\binom{2}{1}\binom{2}{1} + \binom{2}{0}\binom{2}{0}\binom{2}{2}$$
where the terms represent the number of ways of taking one person each from groups $A$ and $B$, one person each from groups $A$ and $C$, one person each from groups $B$ and $C$, and both members of group $C$.
