<s>[INST] 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!
 [/INST] 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.
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