How to prove  $$\frac{3!}{2(n+3)}= \sum_{r=0}^n (-1)^r \frac{ n\choose r}{{r+3}\choose r}$$?  I'm really not getting any nerve to prove this. Can anyone help me in deducing this. Any hint would be appreciated.
Prove that $\displaystyle \sum^{n}_{r=0}(-1)^r\cdot \large\frac{\binom{n}{r}}{\binom{r+3}{r}} = \frac{3!}{2(n+3)}$   $\bf{My\; Try::}$ We can write $$\frac{\binom{n}{r}}{\binom{r+3}{r}} = \frac{n!}{r!\times (n-r)!}\times \frac{r!\times 3!}{(r+3)!} = \frac{3!\times n!}{(n-r)!\times (r+3)!}$$  So our sum is $$ 3!\sum^{n}_{r=0}(-1)^r\frac{n!}{(n-r)!\times (r+3)!} = \frac{3!}{(n+1)(n+2)(n+3)}\sum^{n}_{r=0}(-1)^r\frac{(n+3)!}{(r+3)!\times (n-r)!}$$  So we get Sum $$ \frac{3!}{(n+1)(n+2)(n+3)}\sum^{n}_{r=0}(-1)^r\binom{n+3}{r+3} = -\frac{3!}{(n+1)(n+2)}\sum^{n}_{r=0}(-1)^{r+3}\binom{n+3}{r+3}$$  $$  = -\frac{3!}{(n+1)(n+2)(n+3)}\left[-\binom{n+3}{3}+\binom{n+3}{4}-\binom{n+3}{5}--.......+(-1)^{n+3}\binom{n+3}{n+3}\right]$$  $$=-\frac{3!}{(n+1)(n+2)(n+3)}\left[\binom{n+3}{0}-\binom{n+3}{1}+\binom{n+3}{2}-\binom{n+3}{3}+\binom{n+3}{4}-\binom{n+3}{5}--.......+(-1)^{n+3}\binom{n+3}{n+3}\right]+\frac{3!}{(n+1)(n+2)(n+3)}\left[\binom{n+3}{0}-\binom{n+3}{1}+\binom{n+3}{2}\right]$$  So we get $$ = \frac{3!}{(n+1)(n+2)(n+3)}\left[1-(n+3)+\frac{(n+3)(n+2)}{2}\right] = \frac{3!}{(n+1)(n+2)}\times \frac{n^2+3n+2}{2} = \frac{3!}{(n+1)(n+2)(n+3)}\times \frac{(n+1)(n+2)}{2} = \frac{3!}{2(n+3)}$$  Can we prove it Using Combinatorial way,bcz above method is very lengthy  If yes then plz explain here  Thanks
Why is $\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}$? Show that not all 5 cycles in $A_5$ are conjugate in $A_5$ This is a practice question (not assignment). One thing I am wondering is what this question exactly means. From my understanding, the conjugate of $a \in A_5$ is $gag^{-1} \in A_5$ where $g \in A_5$. I was trying to look at a similar question here on MSE and someone mentioned $(12345) \in A_5$ is not conjuagte in $A_5$. So does that means there exists a $g \in A_5$ such that $g(12345)g^{-1} \notin A_5$ (I am guessing this is incorrect since $A_5$ is a group). I am also away of the property that $g(12345)g^{-1} = (g(1)g(2)g(3)g(4)g(5))$. "Is far from..." - How do I connote the simile, without the literal connotation? 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N: Updating from such a repository can't be done securely, and is therefore disabled by default. N: See apt-secure(8) manpage for repository creation and user configuration details. My English teacher feels very strongly that exact same is redundant and therefore incorrect. I disagree with her.  She feels that exact should be used in place of exact same, but I have rarely heard someone use exact by itself. For example:     I have the exact same shirt   I have the exact shirt   In the same manner, I believe that she shuns exactly the same as well.  What support is there (if any) supporting the use of exact same or exactly the same? What proof can I provide for using exact same when challenged? I can't seem to access instance members of the surrounding class from inside an enum, as I could from inside an inner class. Does that mean enums are static? 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