You are not logged in.
Post a reply
Topic review (newest first)
- bobbym
- 2012-10-30 01:44:34
2!!! = 2 not 1.
- Harold
- 2012-10-30 01:40:30
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
- 2012-10-30 00:57:27
Hi;
That is correct. Your formula looks fine to me. Although there is a better one on that page that anonimnystefy sent you.
- Harold
- 2012-10-30 00:47:18
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
- 2012-10-30 00:09:14
Hi;
That is correct. Now what do you get for 20!!!! ?
- Harold
- 2012-10-29 23:48:16
Well,if I try n=11,k=3,then11!!!=11*8*5*2(as 11 mod 3=2) =880,is it correct?
- bobbym
- 2012-10-29 23:26:57
Hi Harold;
Did you try one? Say for k = 3.
- Harold
- 2012-10-29 22:00:35
Can't this be another definition-
- anonimnystefy
- 2012-10-29 20:12:14
It can be defined recursively:http://en.wikipedia.org/wiki/Factorial#Multifactorials
- Harold
- 2012-10-29 19:51:18
How do you define multifactorial?I know what it is,but how do you define it?I know how to define factorial.