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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.

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  • 2
    $\begingroup$ What does your notation mean. $\endgroup$
    – William Elliot
    Mar 24, 2018 at 10:52
  • $\begingroup$ @JefLaga That is a multiplication $\endgroup$
    – ami_ba
    Mar 24, 2018 at 10:52
  • $\begingroup$ @WilliamElliot C from combinatorical notation $\endgroup$
    – ami_ba
    Mar 24, 2018 at 10:53
  • $\begingroup$ Is the question "solve $5\binom{x}{y}=3\binom{7}{3}$"? $\endgroup$
    – Angina Seng
    Mar 24, 2018 at 11:03
  • 1
    $\begingroup$ Yes, it is....i am new LaTeX user, so I used that notation $\endgroup$
    – ami_ba
    Mar 24, 2018 at 11:07

1 Answer 1

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\begin{align} 5{x \choose y} & = 3{7 \choose 3} \\ & = 3 \cdot \frac{7!}{3!(7-4)!} \\ & = 3 \cdot \frac{7 \cdot 6 \cdot 5 \cdot 4!}{3!\cdot4!} \\ & = 3\cdot \frac{7\cdot 6\cdot 5}{3!} \\ & = 7 \cdot 5 \cdot 3 \\ & = 105 \\ \implies {x \choose y} & = 21 \\ \end{align}

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 $(7,2), (7,5), (21,1),(21,20)$.

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