- Thread starter
- #1

My Try:: Clearly $\displaystyle \binom{n}{r} = 120 \Rightarrow \binom{120}{1} = \binom{120}{119} = 120$

So $(n,r) = (120,1)\;\;,(120,119)$ are positive integer ordered pairs which satisfy the given equation.

Now we will calculate for other positive integer ordered pairs whether it is exists or not.

So $\displaystyle \binom{n}{r} = \frac{n!}{r! \cdot (n-r)!} = 2^3 \times 3 \times 5\Rightarrow \frac{n!}{r! .\cdot (n-r)! \cdot 5} = 2^3 \cdot 3$

So Largest prime factors of $120$ is $5$. So $\displaystyle n\geq 5$

Now for $r$. Here $ 1 \leq r < 119$ and $r \leq \frac{n}{2}$

So my Question is How can I calculate other positive ordered pairs.

So please help me

Thanks