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**Harold****Guest**

How do you define multifactorial?I know what it is,but how do you define it?I know how to define factorial.

**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,922

It can be defined recursively:http://en.wikipedia.org/wiki/Factorial#Multifactorials

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**Harold****Guest**

Can't this be another definition-

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 95,085

Hi Harold;

Did you try one? Say for k = 3.

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**If it ain't broke, fix it until it is.**

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**Harold****Guest**

Well,if I try n=11,k=3,then11!!!=11*8*5*2(as 11 mod 3=2) =880,is it correct?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 95,085

Hi;

That is correct. Now what do you get for 20!!!! ?

**In mathematics, you don't understand things. You just get used to them.**

**If it ain't broke, fix it until it is.**

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**Harold****Guest**

Well,thank you,I love solving problems,now,20!!!!,so n=20 and k=4

20!!!!=20(20-4)(20-4*2)(20-4*3)...(4) [as 20 mod 4=0]=20*16*12*8*4=122880.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 95,085

Hi;

That is correct. Your formula looks fine to me. Although there is a better one on that page that anonimnystefy sent you.

**In mathematics, you don't understand things. You just get used to them.**

**If it ain't broke, fix it until it is.**

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**Harold****Guest**

But,I am confused,if I do 11!!! by the method in wikipedia,I get 11!!!=11(8!!!)=11(8(5!!!))=11(8(5(2!!!)))=11(8(5(1))) [as 0<2<3] =440

and by my method,I get 11!!!=11(11-3)(11-2*3)...(2)=11*8*5*2=880

which is correct ??

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 95,085

2!!! = 2 not 1.

**In mathematics, you don't understand things. You just get used to them.**

**If it ain't broke, fix it until it is.**

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