|
| physics chemistry maths science forums |
|
|
|
| |
|
|
| Author |
Message |
![[Post New]](/templates/default/images/icon_minipost_new.gif) 11 Mar 2008 19:26:19 IST
|
|
|
Prove that (n!)! is divisible by (n!)(n-1)!
|
" Always remember money isn't everything but make sure you have made a lot of it before talking such nonsense!"
- Bill Gates |
|
|
|
|
|
|
|
The basic thing is that the product of any n consecutive numbers is divisible by n!
(n!)! consists of n! numbers = n (n-1)! consecutive numbers.
That is, there are (n-1)! sets of n consecutive numbers, each of which is divisible by n!
Hence (n!)! is divisible by (n!)(n-1)!
|
this reply: 5 points
(with 1 
in 1 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
11 Mar 2008 19:41:28 IST
|
|
|
Since the product of n consecutive int is divisible by n!
Now (n!)! can b written as
(N!)!=1,2,3?.n ---?divisible by n!
(n+1),(n+2)?.2n ---?divisible by n!
(2n+1),(2n+2)?.3n ---?divisible by n!
.
.
.
.
(((n-1)!-1)n+1),,(((n-1)!-1)n+2)?((n-1)!)n ---?divisible by n!
Since we have (n-1)! Rows and each row divisible by n!
Hence proven
How abt that>>>>>>>>>>>
|
The Best remains the BEST |
this reply: 5 points
(with 1
in 1 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
|
|
|
|
|
|
|
Moderation Team
![[Up]](/templates/default/images/icon_up.gif "[Up]"){kind=icon}
