# Thread: integer ordered pair in combination

1. ## integer ordered pair in combination

$Total number of positive integer ordered pair of \binom{a}{b} = 120$

$Using \binom{a}{b} = 120 = \binom{120}{1} = \binom{120}{119}. So (a,b) = (120,1)\;,(120,119)$

$And \binom{a}{b} is \max, when b=\frac{a}{2} or b=\frac{a+1}{2}$

$So must have b\leq \frac{a}{2} or b \leq \frac{a+1}{2}$

$Now How could i get bound on a and b, Thanks$

2. ## Re: integer ordered pair in combination

Originally Posted by juantheron
$Total number of positive integer ordered pair of \binom{a}{b} = 120$

$Using \binom{a}{b} = 120 = \binom{120}{1} = \binom{120}{119}. So (a,b) = (120,1)\;,(120,119)$

$And \binom{a}{b} is \max, when b=\frac{a}{2} or b=\frac{a+1}{2}$

$So must have b\leq \frac{a}{2} or b \leq \frac{a+1}{2}$

$Now How could i get bound on a and b, Thanks$
It isn't exactly clear what you are asking. Nor is your reasoning coherent in places.

Reasoning (although on close inspection, doesn't look like that is what is the question
$Let S be a set \left\{\binom{a}{b}: (a,b)\in\mathbb{Z}\times \mathbb{Z}\right\} whose size is 120$,
$\noindent Observe that the set has 120 members, and using the property \binom{a}{b} = \binom{a}{120-b} \\ using the argument you have given, we know that S is not empty.\\ If that is the question, then I think it needs more thought, the provided reasoning$

Second interpretation (which I think is what you are getting at)
$Let S be a set \left\{\binom{a}{b}=120: (a,b)\in\mathbb{Z}\times \mathbb{Z}\right\}\binom{a}{b}=120 \\ we know that (120,1) and (120,119) are in the set. So there are at least 2 members. \binom{a}{b}=\frac{a!}{(a-b)!\cdot b!}\\[1em]$
$Let us consider b\leq60 \\ Observe that for all b in the restriction of b\leq \left \lceil{\frac{a}{2}}\right \rceil that \binom{a}{b} > \binom{a}{b-1} and \binom{a+1}{b} > \binom{a}{b}$

$\noindentb=1, a=120 \\ b=2, a has 1 solution (using definition of Triangular Numbers) of 16. So consider only a<16$
$\noindent b=3, a=10 , using formula \frac{n(n-1)(n-2)}{6}. So consider only 8
(Have a think why don't we need to consider a=7?)

$\noindent b=4, a has no solution, using formula \frac{n(n-1)(n-2)(n-3)}{24}. This can be checked by putting either 8 or 9$

This is sufficient. (Why?). So we have (120,1); (120,119); (16,2); (16,14); (10,3); (10,7). So answer required is 6.
(excuse the informal/incompleteness of this)

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