# permutations question #2

• May 14th 2009, 08:55 PM
brentwoodbc
permutations question #2
I thought I had this question solved and I sort of did bu I got only half of the answer.
Solve ....
6Pn=720

so what I did was

6!/(6-n)!=720
6!/720=(6-n)! ****Also why cant you cancel the ! here, I mean why cant you divide it out here?
6x5x4x3x2x1/720=6-n
720/720=6-n
n=6-1
n=5

but the answer that is given is 5 or 6? where is this 6 coming from?
• May 14th 2009, 10:04 PM
TwistedOne151
You can't divide it out there because the factorial is an operation, not a variable. You got to $\displaystyle \frac{6!}{720}=(6-n)!$, and then found that 6!=720, right? So then $\displaystyle (6-n)!=1$. You got the case for 1!=1, so that n=5 is a solution, as (6-5)!=1!=1. But there is another solution, as you noted. Here is a hint: what is 0! ("zero factorial")?

--Kevin C.
• May 14th 2009, 10:25 PM
brentwoodbc
0!=1 but I dont see how that is applied to this question, and if its an operation not a variable why can you cancel it later.
• May 14th 2009, 10:34 PM
brentwoodbc
and what I mean by canceling later is my teacher said to do this

example

10Pn=90
10!/(10-n)!=90
10!=90(10-n)! *****I want to know why you cant cancel here.
10!/90=(10-n)!
8!=(10-n)!****but you can cancel here
8=10-n
n=2

but if you canceled at 10!=90(10-n)! you have n= -.888888

I dont see why my first example has 2 solutions and my second has one.
• May 14th 2009, 11:49 PM
TwistedOne151
Because you had (6-n)!=1. 1!=1, and as you noted, 0!=1, which means that 6-n can be 1 or 0, and thus n=5 or 6.

--Kevin C.
• May 15th 2009, 10:39 AM
brentwoodbc
Quote:

Originally Posted by TwistedOne151
Because you had (6-n)!=1. 1!=1, and as you noted, 0!=1, which means that 6-n can be 1 or 0, and thus n=5 or 6.

--Kevin C.

ya thanks I thought about it and also 7*6*5*4*3*2 when you cancel is =1!, =7*6*5*2*1 when you cancel its = 0! then when you cancel the factorials your left with 1/0. thanks.