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Combinatorics

Can anyone help me find a literature mentioning the following formula. I formulated it a long time ago and I am not sure if someone had thought of it before me.

\(_kC_r=\sum_{n=1}^{k-r+1} (_{k-n}C_{r-1})\) for \(k\geq r\geq 2\)

Note by Roman Frago
1 year, 10 months ago

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@Kishlaya Jaiswal

Have you read this formula ever before ? I'm pretty sure only you can help Mr. Frago out :) Azhaghu Roopesh M · 1 year, 10 months ago

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@Azhaghu Roopesh M Yep, it's indeed the Hockey-Stick Formula

I've mentioned clearly about it in my article, so you may wish to read it.

Yeah, now I've got a nice topic for a new wiki. Kishlaya Jaiswal · 1 year, 10 months ago

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@Kishlaya Jaiswal It's different from the Hockey-Stick formula. Roman Frago · 1 year, 10 months ago

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@Roman Frago Nope ,it is indeed (just a little bit of manipulations and sir, you'll clearly see the identity). Allow me to explain

\[\begin{eqnarray} \sum_{n=1}^{k-r+1} {k-n \choose r-1} & = & {k-1 \choose r-1} + {k-2 \choose r-1} + \ldots + {r-1 \choose r-1} \end{eqnarray}\]

Setting \(r-1=t\), and then writing the reverse expression, gives \[\begin{eqnarray} \sum_{n=1}^{k-t} {k-n \choose t} & = & \sum_{n=0}^{k-t-1} {t+n-1 \choose t} \\ & = & {t \choose t} + {t+1 \choose t} + \ldots + {k-2 \choose t} + {k-1 \choose t} \\ & = & {k \choose t+1} \\ & = & {k \choose r} \end{eqnarray}\]

which follows directly from Hockey Stick Identity

Thanks. Kishlaya Jaiswal · 1 year, 10 months ago

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@Kishlaya Jaiswal Yes it is . I had a wrong reference .(http://www.artofproblemsolving.com/Wiki/index.php/Pascal%27s_triangle)

But this one clearly shows that it is the Hockey-Stick:http://www.artofproblemsolving.com/Wiki/index.php/Combinatorialidentity#Hockey-StickIdentity.

Thank you very much.

EDIT: By the way, I thought of it looking into a pentagram. Roman Frago · 1 year, 10 months ago

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@Kishlaya Jaiswal You can even workout the complete proof of your above stated identity just by considering the expansions of

\[\left(\frac{1}{1-x}\right)^{r+1} = \left(\frac{1}{1-x}\right)^{r}\left(\frac{1}{1-x}\right)\]

and then comparing the coefficients of \(x^{k-r}\) Kishlaya Jaiswal · 1 year, 10 months ago

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@Kishlaya Jaiswal I am not looking for proof. I am trying to see if anyone has formulated it before me.

EDIT: I appreciate you looking into this. Roman Frago · 1 year, 10 months ago

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@Kishlaya Jaiswal Have you read my comment at the bottom of this page ? No hurry, but can you please respond , so that I'll get a better idea of how to proceed

Thanks :) Azhaghu Roopesh M · 1 year, 10 months ago

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@Azhaghu Roopesh M Yep I've already read it. But actually, I'll take some to reply because I haven't tried out your chemistry problems yet. I will surely give them a try tomorrow. Kishlaya Jaiswal · 1 year, 10 months ago

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@Kishlaya Jaiswal I see , so that's why I was able to link it with you :)

I had just used it once as an identity , more like just a formula . I am wondering what else can be there to it !!

I'll be eagerly waiting for your wiki :)

P.S. Congrats on your getting 200 Friends !! Azhaghu Roopesh M · 1 year, 10 months ago

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@Azhaghu Roopesh M @Kishlaya Jaiswal,@Ronak Agarwal ,@Pranjal Jain ,@Deepanshu Gupta ,@megh choksi ,@Sudeep Salgia ,@Sandeep Bhardwaj sir ,@Raghav Vaidyanathan ,@Akshay Bodhare and everyone on Brilliant .

Did you guys see the Chemistry questions that I have posted recently ? If so , can you give me a feedback on it's level and pretty much anything else you would like to tell me ?

I just wanted Chemistry to be as popular on Brilliant as are Maths and Physics .

Please give a honest feedback , I won't mind if you criticize me .

Last but not the least , the king of Brilliant @Calvin Lin sir , even your comment will be invaluable :) Azhaghu Roopesh M · 1 year, 10 months ago

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@Azhaghu Roopesh M I think posting organic reactions won't be good idea. Though I just saw a glimpse of it and I don't think I saw any reaction in your set except that Qualitative Analysis. Pranjal Jain · 1 year, 10 months ago

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@Pranjal Jain Ok , so I won't be posting any more Organic Chem questions .

I understand why you have said that , thanks for replying :) Azhaghu Roopesh M · 1 year, 10 months ago

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