Stewart's Hoard: Magic Circle

Arrange the numbers 1, 2, 3, 4, 5, and 6 in the small circles so that the sum of the numbers on each large circle is fourteen.

First, notice that each small circle is intersected by two large circles.  If I give each circle its own color, we have the following.

For each small circle we have three options,
Red/Green
Green/Blue
Blue/Red
and each of these appears twice in the diagram





In order to solve this problem I first began with C(6;4) to find all the four-number combinations of the numbers 1-6.  For those of you who don't know, C(6;4) = $\dfrac{6!}{4!(6-4)!} = 15$. 
Here are the fifteen combinations:

1234     1235     1236     1245     1246    
1256     1345     1346     1356     1456
2345     2346     2356     2456     3456

Now find those that have a sum of fourteen:

1234     1235     1236     1245     1246    
1256     1345     1346     1356     1456
2345     2346     2356     2456     3456

There are three combinations and three circles and now our task is complete.  As you can see, each of the combinations shares two digits with another combination, and the other two digits with the remaining combination.  This allows us to fulfill the intersections of the circles.

Source of Problem: Professor Stewart's Hoard of Mathematical Treasures by Ian Stewart