- #1

juantheron

- 21

- 1

My Solution:: 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