-1
$\begingroup$

Symbolic Representation of the Equality

$${}^{2n}C_{(3r-1)} = {}^{2n}C_{(r+1)}$$

See the image and provide me with the conditions in which this is possible! I have searched the web but couldn't really find anything about it, maybe i was searching for wrong keywords.

  • 0
    Do you mean $$\left(\begin{array}{c}2n\\ 3r-1\end{array}\right)=\left(\begin{array}{c}2n\\ r+1\end{array}\right)$$ ? – 2017-01-27
  • 0
    Yeah! exactly!! – 2017-01-27
  • 0
    almost done ....look below – 2017-01-27
  • 0
    Gee, thanks for the help! – 2017-01-27

1 Answers 1

4

Hint:$$\left(\begin{array}{n}n\\ k_1\end{array}\right)=\left(\begin{array}{c}n\\ k_2\end{array}\right)\Rightarrow \begin{cases}k_1=k_2\\k_1+k_2=n\end{cases}$$so $$\left(\begin{array}{c}2n\\ 3r-1\end{array}\right)=\left(\begin{array}{c}2n\\ r+1\end{array}\right)\\ $$ $$\left(\begin{array}{n}2n\\ 3r-1\end{array}\right)= \left(\begin{array}{c}2n\\ r+1\end{array}\right)\Rightarrow \begin{cases}3r-1=r+1\\(3r-1)+(r+1)=2n\end{cases}\\\begin{cases}3r-1=r+1 (*) \\(3r-1)+(r+1)=2n (**)\end{cases}$$ Note that $\left(\begin{array}{c}n\\ k\end{array}\right)\to 0\leq k\leq n$ $$(*) \to 3r-1=r+1 \to r=1 \space \to \left(\begin{array}{c}2n\\ 2\end{array}\right)=\left(\begin{array}{c}2n\\ 2\end{array}\right) \checkmark$$ $$(**) \to 4r=2n \to 2r=n \to \\\left(\begin{array}{c}4r\\ 3r-1\end{array}\right)=\left(\begin{array}{c}4r\\ r+1\end{array}\right) \to \begin{cases}0\leq 3r-1\leq 4r\\0\leq r+1\leq 4r\end{cases}$$