Find the number of ordered pairs $(x,y)$ such that $5\cdot{}^xC_y=3\cdot{}^7C_3$

Find the number of ordered pairs $(x,y)$ such that $5\cdot{}^xC_y=3\cdot{}^7C_3$

I am very new to number theory and combinatorics, so I could not proceed much. I tried to break ${} ^7C_3$ in the factorial form and then tried to simply, but it was not of any help.

So, any hint or answer will be appreciated.

• What does your notation mean. – William Elliot Mar 24 '18 at 10:52
• @JefLaga That is a multiplication – ami_ba Mar 24 '18 at 10:52
• @WilliamElliot C from combinatorical notation – ami_ba Mar 24 '18 at 10:53
• Is the question "solve $5\binom{x}{y}=3\binom{7}{3}$"? – Lord Shark the Unknown Mar 24 '18 at 11:03
• Yes, it is....i am new LaTeX user, so I used that notation – ami_ba Mar 24 '18 at 11:07

We know that $x$ is at most $21$, because the values are increasing as $x$ increases. Obvious solutions are $(21,1)$ and $(21,20)$. Looking at Pascal's Triangle, we see that $(7,2)$ and $(7,5)$ are answers as well.
Hence there are $4$ solutions which are $(7,2), (7,5), (21,1),(21,20)$.