Combinations of 5 integers from 1 to 100 such that differences between the sorted integers of each combination is at least 5 but not more than 10? For example , I am trying to count combinations like [1,6,14,21,27] because the minimum difference between two sequential integers in the combination is 5 and the maximum distance is 8, but I don't want to count combinations like [2,4,9,16,25] because the smallest difference between the sorted numbers is 2 (too small) or combinations like [5,30,35,41,48] because the largest distance between sorted numbers is 25 (too large).
I know that the total number of ways to select 5 numbers from 100 is "100 choose 5" = 75,287,520 so the answer must be less than this.  
Is there a general formula or algorithm for this type of problem?

Choose the first integer
$$\sum_{q=1}^{100} z^q = z \sum_{q=0}^{99} z^q
= z \frac{1-z^{100}}{1-z}$$
Choose four gaps at least five and not more than ten:
$$(z^5+z^6+\cdots+z^{10})^4
= (z^5 (1+\cdots+z^5))^4
= z^{20} \frac{(1-z^6)^4}{(1-z)^4}.$$
Extract those values that terminate in at most one hundred:
$$[z^{100}] \frac{1}{1-z}
z^{20} \frac{(1-z^6)^4}{(1-z)^4} 
z \frac{1-z^{100}}{1-z}
\\ = [z^{79}] \frac{1}{(1-z)^6}
(1-z^6)^4 (1-z^{100}).$$
Now the multiple of $z^{100}$ does not contribute to $[z^{79}]$
and we get
$$[z^{79}] \frac{1}{(1-z)^6} (1-z^6)^4
= [z^{79}] \frac{1-4z^6+6z^{12}-4z^{18}+z^{24}}{(1-z)^6}$$
This is
$$[z^{79}] \frac{1}{(1-z)^6}
- 4 [z^{73}] \frac{1}{(1-z)^6}
\\ + 6 [z^{67}]  \frac{1}{(1-z)^6}
- 4 [z^{61}]  \frac{1}{(1-z)^6}
+ [z^{55}]  \frac{1}{(1-z)^6}
\\ = {79+5\choose 5}
- 4 {73+5\choose 5}
\\ + 6 {67+5\choose 5}
- 4 {61+5\choose 5}
+ {55+5\choose 5}
\\ = 90720.$$
