What number of positive integers satisfying the given inequality? What number of positive integers satisfying the given inequality : 
              C(n + 1,n – 2) – C(n + 1,n – 1)<= 100

What I did so far :
C(n + 1,n – 2) – C(n + 1,n – 1) <= 100 
=> C(n + 1,3)  – C(n + 1,2) <= 100 
=> n(n+1)(n-4) <= 600

Now to derive after this ? 
PS: This is not my homework, It originated my test paper without proper explanation of answers.
EDIT:The time constraint is only a mint so please suggest shortest possible method.
 A: To solve $n(n+1)(n-4)\le 600$ algebraically, expand the left side and subtract 600 from both sides, so that you're comparing a polynomial to 0, find the zeros of the polynomial—which divide the number line into regions—and test a value in each region to see which regions satisfy the inequality.
Simpler, especially since you're only looking for positive integer solutions, would be to graph $y=x(x+1)(x-4)-600$ and look for where it is below the x-axis.

edit: Under very tight time constraints and without a calculator/computer, I'd have approached the original problem differently.  The original statement  ${n+1\choose n–2}–{n+1\choose n–1}\le 100$  (as well as the restatement ${n+1\choose 3}–{n+1\choose 2}\le 100$) can be interpreted as a statement about the (n+1)st row of Pascal's triangle: that the difference between the numbers 2 and 3 positions in from the edge is less than or equal to 100.  It should be clear that the difference between adjacent entries in a row of Pascal's triangle is increasing, so it is sufficient to write successive rows until the statement fails:

The bold/darkest numbers are the relevant ones:
n=1: $0-1\le 100$;
n=2: $1-3\le 100$;
n=3: $4-6\le 100$;
n=4: $10-10\le 100$;
n=5: $20-15\le 100$;
n=6: $35-21\le 100$;
n=7: $56-28\le 100$;
n=8: $84-36\le 100$;
n=9: $120-45\le 100$;
n=10: $165-55>100$.
Also, note that the lightest numbers, below and to the left of the relevant numbers, are not used at all, so you don't actually need to compute them.
A: $n(n+1)(n-4) \le 600$
It is just a matter of trial and error, as $n(n+1)(n-4)$ is increasing for $n \ge 4$. For $ 0 < n < 4$ it is negative.
For $n=10$, the value is $660 > 600$ and for $n=9$ the value is $450$.
Thus the number is $9$ got by $1 \le n \le 9$.
