Number of Interesting Quadruples 
Define an ordered quadruple of integers $(a, b, c, d)$ as interesting if $1 \le a<b<c<d \le 10$, and a+d>b+c. How many interesting ordered quadruples are there?

This is a bit of trouble here actually, I am to use $a + d \gt b + c$ as a constraint. 
Without any restrictions (no $a + d \gt b + c$) there are: $\binom{10}{4} = 210$ possible values for $a, b, c, d$. 
We could have three cases: $a + d \gt b + c$ or $a + d < b + c$ or $a + d = b + c$.
We need to take out $a + d = b + c$ cases first: It is possible that: 
$a + b  = \{1 + \sum_{k=4}^{10}k, 2 + \sum_{k=5}^{10}, 3 + \sum_{k=6}^{10}, 4 + \sum_{k=7}^{10}k, ..., 7 + 10 \}$
Total (incl. overcounting): $7 + 6 + 5 + 4 + 3 + 2 + 1 = 28$ possible. 
But I think I have messed the whole problem.
Hints Please!

Here is my solution: 

Restate the inequality as $b-a < d-c$ so that the range of numbers for a and b is strictly less than for d and c.
Now, there are $\binom{10}{4}$ total possible choices of the 4 different integers in the given range. $\binom{8}{2}$ of those choices are such that the pairs (a,b) and (d,c) each have a range of 2 (i.e. when (a,b) are adjacent and (c,d) are adjacent), $\binom{6}{2}$ are such that (a,b) and (c,d) each have a range of 3, $\binom{4}{2}$ where the pairs have a range of 4 and $\binom{2}{2}$ where each have a range of 5.
By subtracting these choices from our total we are left with those choices where $b-a\ne d-c$ and by symmetry exactly $1/2$ of these are of the desired type, hence our require count is: $$\frac{1}{2}\left(\dbinom{10}{4}-\dbinom{8}{2}-\dbinom{6}{2}-\dbinom{4}{2}-\dbinom{2}{2}\right)$$
This gives us:
$$\frac{1}{2}\left(210 - 28 - 15 - 6 - 1\right) = \frac{160}{2} = 80$$
