Solve for n?

Question

n! - !n = n

^_^

Answer 1

For those who do not know: !n is called the subfactorial of n, the number of derangements of n elements.

The equation n! - !n = n is only satisfied when n=0 or 1.

Answer 2

n=0

Answer 3

Here is a proof of Goitre19's statement using a recurrence relation found by Euler, namely:

!n = n*(!(n-1)) + (-1)^n

and the obvious recurrence relation for factorial, namely:

n! = n*(n-1)!

First, if n = 0, then as 0! = 1 = !0, we have 0! - !0 = 1 - 1 = 0.

Now assume that n ≠ 0. Then

n = n! - !n = n*(n-1)! - n*(!(n-1)) - (-1)^n = n*((n-1)! - !(n-1)) - (-1)^n.

Rewriting we get n*((n-1)! - !(n-1)) = n + (-1)^n. The right side of this equation is either n - 1 or n + 1, and so we have either n divides n - 1 or n divides n + 1. In either case, the only possible solution is n = 1.

∴ n = 0 and n = 1 are the only solutions to n! - !n = n.

Answer 4

n! = n.(n-1)!..
so puttin this value in the left hand side.. we get..
n(n-1)! - !n = n
taking n common,
n { (n-1)! - !n } = n...

the n will then cancell...and i think the answer wud be 0..
check if the ques is correct..i doubt..

Answer 5

n! we already knew, but - !n may be 1!n
which means :
n! - n = n
n! = 2n
n* !(n-1) = 2n
!(n-1) = 2
n-1 = 2
n = 3

Answer 6

This isn't a serious question is... oh well, 2 points for me.

Answer 7

this in factorinal n - not n = n.....

n! - !n = n

cancelling the n's we have

! - ! = 1
0 =1

hope that's fine.

Answer 8

n is simply a numer which i dont know LOL lmao

Answer 9

n=n

Answer 10

there must be a mistake