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 \(a>b\).

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 \(a>b\).

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TopNewestCan 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 · 2 years, 2 months agoLog in to reply

– Abhishek Sharma · 2 years, 2 months ago

My bad. I have corrected it now.Log in to reply