Counting combinations with a restriction of the form "either ... or ..., but not both" The following is the problem that I am dealing with.

There are 9 people in a class and 4 of them is randomly chosen to form a committee.  Jack and Nick are 2 of the 9 people in the class.  How many committees can be formed which Jack or Nick are in it, but not both?

Supposedly the answer is $70$, but I have trouble finding the answer and this is what I have so far.
Using the counting principle, there are 4 seats available for the committee so if Jack or Nick are in it I would calculate
$$2*8*7*6$$
Since this is a group and order does not matter I divided by $4!$.
Now I am thinking of the Venn diagram where I am doing 
$$J+N-2(J \cap N)$$
so now I need to think about the number of committees with Jack and Nick in it as
$$1*1*7*6$$
where 1 and 1 represents Jack and Nick, and since there are only 7 people left the rest of the people should be $7 * 6$.
However, I noticed that if I divide this by the number of order of 4 people it gives me a fraction.
Can someone explain me why my method is not working and how you would solve this problem?
 A: You're miscounting the number of committees with Jack or Nick on them. To illustrate what is going wrong, there, let's look at a simpler example. Suppose we have a class of only three people--Abe, Jack, and Nick--and want to make a committee of two of the members of the class. Clearly, every such committee will have Jack or Nick on it, but if we use a parallel argument to your counting method, this would be $2\cdot2=4,$ which we would then divide by $2!=2$ since the order doesn't matter, to get $2.$ But three committees are possible!
A better way to count the number of committees with Jack or Nick on it is to first find the number of committees--namely $\frac{9\cdot8\cdot7\cdot6}{4!}=126$--then subtract the number of committees that don't have Jack or Nick on them--namely $\frac{7\cdot6\cdot5\cdot 4}{4!}=35$--to find that there are $91$ committees with Jack or Nick on them. Once that's done, we can subtract the number of committees with Jack and Nick on them--namely $\frac{7\cdot6}{2!}=21$ (we only have to choose two for these)--which leaves us with $70$ committees with Jack or Nick on them, but not both.
A: You have 9 people, two of which are Jack and Nick. Since they must be in your committees, your problem is how many ways you have of selecting 3 people out of the remaining 7, then add either Jack or Nick to the committee. Therefore:
$2 · \frac{7·6·5}{3!} = 70$
A: My way of thinking is: 
First consider if only Jack is in the committee, so we need to choose another 3 people out of 7 (Jack is chosen and Nick cannot be chosen), which is $\binom{7}{3}=35$. 
Same goes if only Nick is in the committee. So the number of committees can be formed which Jack or Nick are in it, but not both, is $2\binom{7}{3} = 70$.
