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Prove that $\displaystyle \prod _{ r=0 }^{ a } \binom{a}{r} =\left(\prod _{ r=0 }^{ b } \binom{b}{r} \right)\left(\prod _{ r=1 }^{ a-b }{ \frac { { (b+r) }^{ b+r-1 } }{ (b+r-1)! } }\right )$ , where $a>b$ .

I had previously posted this as a question which hadn't got much response so I am now posting this as a note.

Try more problems here.

#Combinatorics #Proofs #Awesome

Easy Math Editor

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

Can I mention really quickly that in this problem, you didn't specify that the brackets meant the floor function? To avoid ambiguity, write $\lfloor x\rfloor$ (code: \lfloor x\rfloor) instead of $[x]$ . (P.S. How do you do a private message here?)

Akiva Weinberger - 6 years, 1 month ago

My bad. I have corrected it now.

Abhishek Sharma - 6 years, 1 month ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$