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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.

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For a start, there is a small slip in what you have so far, (n-4) should be (2n-5), I think. – Oscar Cunningham Oct 10 '10 at 17:58
@Oscar Cunningham: I didn't work out the algebra, but the solution set in positive integers for the expression as given is the same as the original problem, but using (2n-5) in place of (n-4) does not give the same solution set as the original problem. – Isaac Oct 10 '10 at 18:01
Yup, I'm an idiot. 3! is not 3 – Oscar Cunningham Oct 10 '10 at 18:27
up vote 2 down vote accepted

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:

Pascal's triangle, through the 11th row

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.

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Ehh,.. I am suppose to solve to the whole thing within a mint, so do you think graph is a good idea ? :) – Quixotic Oct 10 '10 at 18:02
The polynomial $x(x+1)(x-4)$ is increasing except for a little dip between roughly -1 and 3 where the value is nowhere near 600, so it's just a matter of finding the smallest integer which gives a value larger than 600. ($x=9$ gives 450, $x=10$ gives 660; hence $n=1,\ldots,9$ are the positive integer solutions.) – Hans Lundmark Oct 10 '10 at 18:07
@Debanjan: You never mentioned time constraints or what tools you were allowed. Given a computer or graphing calculator, graphing is likely to be the fastest and using a table of values the second-fastest. By hand under very short time pressure, I'd have approached the problem entirely differently: keep it in combinatoric form, start writing rows of Pascal's triangle, and look for where the difference between those adjacent entries in Pascal's triangle stops being less than or equal to 100. – Isaac Oct 10 '10 at 18:46
well, could you please explain your approach ? I am very inquisitive :) – Quixotic Oct 10 '10 at 18:56
@Debanjan: I've edited my answer to include this approach. – Isaac Oct 10 '10 at 19:20

$n(n+1)(n-4) \le 600$

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 $1 \le n \le 9$.

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+1 for suggesting a feasible (considering the time constraint) method. – Quixotic Oct 10 '10 at 18:15

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