# Factorial Notation

• Aug 17th 2011, 12:10 PM
momofmaxncoop
Factorial Notation
I keep having a hard time solving this factorial notation question. I know the steps but I'm having a brain fart of what the correct answer be to input into the next step.

(n+2)! =110
(n)!

(n+2)x(n+1)x(n)(n-1)(n-2)....x1
(n)x(n-1)x(n-2)....x1

(n+2)(n+1)

This is where I get confused.

would the correct answer be n(squared)+3 = 110??

Any help would be gratefully appreciated.
• Aug 17th 2011, 12:23 PM
Siron
Re: Factorial Notation
What do you have to do? Find a $n$ wherefore $(n+2)!=110$? or? ...
• Aug 17th 2011, 12:23 PM
TheChaz
Re: Factorial Notation
Is the question
(n + 2)! = 120 ???????
The result should come "by inspection".

Also, you should know a few factorials.
0! = 1
1! = 1
2! = 2
3! = 6
4! = 24
5! = 120
etc
So 5! = (n + 2)!
5 = n + 2
n = 3

This is, of course, if the problem is 120 (not 110)!
• Aug 17th 2011, 12:32 PM
Also sprach Zarathustra
Re: Factorial Notation
Quote:

Originally Posted by TheChaz
Is the question
(n + 2)! = 120 ???????
The result should come "by inspection".

Also, you should know a few factorials.
0! = 1
1! = 1
2! = 2
3! = 6
4! = 24
5! = 120
etc
So 5! = (n + 2)!
5 = n + 2
n = 3

This is, of course, if the problem is 120 (not 110)!

&#40;x&#43;2&#41; facterial&#61;110 solve for x - Wolfram|Alpha
• Aug 17th 2011, 12:35 PM
momofmaxncoop
Re: Factorial Notation
I meant the first equation to say : (n+2)! / (n!) = 110
• Aug 17th 2011, 12:40 PM
Also sprach Zarathustra
Re: Factorial Notation
Quote:

Originally Posted by momofmaxncoop
I meant the first equation to say : (n+2)! / (n!) = 110

$\frac{(n+2)!}{n!}=\frac{(n+2)(n+1)...2\cdot 1}{n(n-1)...2\cdot 1}=(n+2)(n+1)$

$(n+2)(n+1)=110$
• Aug 17th 2011, 12:40 PM
Plato
Re: Factorial Notation
Quote:

Originally Posted by momofmaxncoop
I keep having a hard time solving this factorial notation question. I know the steps but I'm having a brain fart of what the correct answer be to input into the next step.

(n+2)! =110
(n)!

Surely the question means $\frac{(n+2)!}{n!}=110$.
That is
\begin{align*} \frac{(n+2)!}{n!}&= (n+2)(n+1)\\ (n+2)(n+1)&=110 \\ n^2+3n-108&=0 \end{align*}
• Aug 17th 2011, 12:51 PM
momofmaxncoop
Re: Factorial Notation
Where are you getting the -108??? shouldn't that be -110??
• Aug 17th 2011, 12:59 PM
Plato
Re: Factorial Notation
Quote:

Originally Posted by momofmaxncoop
Where are you getting the -108??? shouldn't that be -110??

\begin{align*}(n+2)(n+1) &=110\\n^2+3n+2 &= 110\\ n^2+3n+2-110&= 0\\ n^2+3n-108&= 0 \end{align*}
• Aug 24th 2011, 03:01 AM
downthesun01
Re: Factorial Notation
Can someone explain to me what you guys are doing in this problem? Are you just solving for n? If so then why isn't the answer just intuitively n=9?

Nevermind, I think I get it now. After solving the polynomial I get n=9, 12

But what's with the 12? It works with the polynomial but obviously not with the equation using factorials.

Just curious
• Aug 24th 2011, 03:48 AM
CaptainBlack
Re: Factorial Notation
Quote:

Originally Posted by downthesun01
Can someone explain to me what you guys are doing in this problem? Are you just solving for n? If so then why isn't the answer just intuitively n=9?

Nevermind, I think I get it now. After solving the polynomial I get n=9, 12

But what's with the 12? It works with the polynomial but obviously not with the equation using factorials.

Just curious

The two solution are n=9, -12 which are both solutions to the quadratic but the second does not solve the original problem since n must be >=0.

So when you get to the quadratic you are looking for a non-negative n which satisfies the quadratic. The solution to that problem is n=9.

CB