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!
 A: 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 $a+d \gt b+c$
A: Revised answer using "stars and bars"
I will write the constraint as $d-c > b-a$ 
We know the first digit must be 1 or higher and the last digit must be 10 or lower, so we can conceive the problem to placing 10 +'s in 5 bins (the last to take up the slack from 10)
To cater to the inequalities and the constraint, we need to pre-place some $+'s$
|+ | + | + | ++ |(+++++)| meaning 1, 2, 3, 5, in the 1st 4 bins & slack of 5 in the fifth bin. ++ has been placed in the 4th bin so that at the start, $d-c > a-b$
We can redistribute the 5 +s in 3 patterns to maintain the constraint
0-0, 1-1, or 2-2 +'s added in b and d, and the rest (5, 3 and 1 respectively) distributed freely excluding bin b, i.e. in 4 bins
Applying the "stars and bars" formula,
$$\text{# of ways = }{8\choose 3}+{6\choose 3} + {4\choose 3} = 80$$ 
A: Let $t_1=a,\; t_2=b-a,\; t_3=c-b,\; t_4=d-c$; 
so $t_1+t_2+t_3+t_4=d$ and $a+d>b+c\implies
t_1+t_2+t_3+t_4\le10$ and $t_4>t_2$.
Now let $t_5=t_4-t_2$ and $t_6=10-(t_1+t_2+t_3+t_4)$ to get
$\hspace{.3 in}2t_2+t_1+t_3+t_5+t_6=10$ with $t_i\ge1$ for $i<6$ and $t_6\ge0$.
Letting $y_i=t_i-1$ for $i<6$ and $y_6=t_6$ gives
$\hspace{.3 in}2y_2+y_1+y_3+y_5+y_6=5$ with $y_i\ge0$ for each $i$, so taking $y_2=0, y_2=1, \text{ and }y_2=2$ 
gives a total of $\displaystyle\binom{8}{3}+\binom{6}{3}+\binom{4}{3}=80$ solutions.
A: 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:
$$\frac{1}{2}\left(210 - 28 - 15 - 6 - 1\right) = \frac{160}{2} = 80$$
