Solving Factorial Notation Questions: An Explanation

[funktion]

Hey, I was wondering if someone could help me with a specific type of question that I can't seem to understand without an answer key. Anyway, it's rewriting expressions with factorial notation so that they no longer have factorial symbols.

Example:

Simplify without using the factorial symbol:
(n-2)!(n+1)!/(n!)²

The answer is: n+1/n(n-1)

What I don't understand is how you come to that conclusion. Can someone explain this to me?

 

[mantito]

well, the key to solution is to notice that n!=(n-2)!*(n-1)*n and (n+1)!=n!*(n+1). also (n!)^2=n!*n!.

 

[dav2008]

What is n! ?

It's n*(n-1)*(n-2)*(n-3)*...*1

What is (n-2)! and (n+1)! ?

If you write all of those out, you'll notice that certain terms cancel.

Edit: Or I guess a more direct approach would be to write n! in terms of (n-2)! like mantito has done.

 

[funktion]

Yeah, I understand what you're saying, but I'm still stuck.

I guess I'll show you my work.

(n-2)!(n+1)!/(n!)²

= (n-2)(n-1)n!(n+1)!/n(n-1)(n-2)!(n!)

= (n+1)!/n!

What I have trouble with is I guess why the answer is (n+1)/n(n-1). Can I not just cancel out the factorial symbol without having to multiply (n-1)?

 

[nrqed]

Yeah, I understand what you're saying, but I'm still stuck.

I guess I'll show you my work.

(n-2)!(n+1)!/(n!)²

= (n-2)(n-1)n!(n+1)!/n(n-1)(n-2)!(n!)

= (n+1)!/n!

What I have trouble with is I guess why the answer is (n+1)/n(n-1). Can I not just cancel out the factorial symbol without having to multiply (n-1)?

I don`t understand your numerator..did you use (n-2)!=(n-2)(n-1) n! ??
That is incorrect!

You just have to write the (n+1)! as (n+1) n! and then write one of the n! of the denominator as n (n-1) (n-2)! and then all the factorials will cancel out leaving you with (n+1) / (n (n-1))

Patrick

 

[funktion]

I don`t understand your numerator..did you use (n-2)!=(n-2)(n-1) n! ??
That is incorrect!

You just have to write the (n+1)! as (n+1) n! and then write one of the n! of the denominator as n (n-1) (n-2)! and then all the factorials will cancel out leaving you with (n+1) / (n (n-1))

Patrick

Thanks a bunch, but one more question: Why was mine incorrect?

 

[lypaza]

Because n! = (n-2)! (n-1) n

 

[VietDao29]

Thanks a bunch, but one more question: Why was mine incorrect?

The numerator in your second step is wrong!

(n-2)!(n+1)!/(n!)²

= (n-2)(n-1)n!(n+1)!/n(n-1)(n-2)!(n!)

= (n+1)!/n!

If it reads:
(n - 2)! n! (n + 1) / (n (n - 1) (n - 2)! n!), then it's correct.
Note that:
(n - 2)! \neq (n - 2) (n - 1) n!
The LHS can be expanded as:
(n - 2) (n - 3) (n - 4) ... 2 . 1
Whereas the RHS is:
n (n - 1)² (n - 2)² (n - 3) ... 2 . 1
And of course the LHS is not equal the RHS, right?
Can you get this? :)