integer ordered pair in combination (1 Viewer)

juantheron · May 26, 2017

$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$

dan964 · May 26, 2017

juantheron said:
$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}$

$$\noindent$b=1, a=120 $\\$ b=2, a $ has 1 solution (using definition of Triangular Numbers) of $16. $ So consider only $a<16$
$$\noindent $ b=3, a=10 $, using formula $ \frac{n(n-1)(n-2)}{6}. $So consider only $ 8<a<10$
(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)

integer ordered pair in combination (1 Viewer)

juantheron

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dan964

what

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