1. ## 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.

2. ## Re: Factorial Notation

What do you have to do? Find a $\displaystyle n$ wherefore $\displaystyle (n+2)!=110$? or? ...

3. ## 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)!

4. ## Re: Factorial Notation

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

5. ## Re: Factorial Notation

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

6. ## Re: Factorial Notation

Originally Posted by momofmaxncoop
I meant the first equation to say : (n+2)! / (n!) = 110
$\displaystyle \frac{(n+2)!}{n!}=\frac{(n+2)(n+1)...2\cdot 1}{n(n-1)...2\cdot 1}=(n+2)(n+1)$

$\displaystyle (n+2)(n+1)=110$

7. ## Re: Factorial Notation

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 $\displaystyle \frac{(n+2)!}{n!}=110$.
That is
\displaystyle \begin{align*} \frac{(n+2)!}{n!}&= (n+2)(n+1)\\ (n+2)(n+1)&=110 \\ n^2+3n-108&=0 \end{align*}

8. ## Re: Factorial Notation

Where are you getting the -108??? shouldn't that be -110??

9. ## Re: Factorial Notation

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

10. ## 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

11. ## Re: Factorial Notation

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