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#1 2012-10-29 19:51:18

Harold
Guest

Multifactorial

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

#2 2012-10-29 20:12:14

anonimnystefy
Real Member

Online

Re: Multifactorial


The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

#3 2012-10-29 22:00:35

Harold
Guest

Re: Multifactorial

Can't this be another definition-

#4 2012-10-29 23:26:57

bobbym
Administrator

Online

Re: Multifactorial

Hi Harold;

Did you try one? Say for k = 3.


In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

#5 2012-10-29 23:48:16

Harold
Guest

Re: Multifactorial

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

#6 2012-10-30 00:09:14

bobbym
Administrator

Online

Re: Multifactorial

Hi;

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


In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

#7 2012-10-30 00:47:18

Harold
Guest

Re: Multifactorial

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.

#8 2012-10-30 00:57:27

bobbym
Administrator

Online

Re: Multifactorial

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.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

#9 2012-10-30 01:40:30

Harold
Guest

Re: Multifactorial

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 ??

#10 2012-10-30 01:44:34

bobbym
Administrator

Online

Re: Multifactorial

2!!! = 2 not 1.


In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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