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!

For each set $a,b,c,d$ with $a+d \neq b+c$, either $a+d \gt b+c$ or $(10-d)+(10-a) \gt (10-b)+(10-c)$.  Exactly half the selections with $a+d \neq b+c$ will meet your restriction.  
I haven't found a neat way to count the cases $a+d=b+c$.  We can note that there are two ways for two numbers to sum to $5$, two ways to sum to $6$, three ways to sum to $7$ up to five ways to sum to $10$, then decreasing to two ways to sum to $17$.  If you pick two ways out of any of these, you get a case $a+d=b+c$, so there are $4{2\choose 2}+4{3 \choose 2}+4{4 \choose 2}+{5 \choose 2}=4(1+3+6)+10=50$ ways to have $a+d=b+c$ and $(210-50)/2=80$ ways to have $a+d \gt b+c$
