Evaluating Fractions with fractorials

[kuahji]

Evaluate each factorial expression

A) (n+2)!/n!

My book doesn't really show how to come up with a solution. After looking in the back for the answer it showed (n+2)(n+1), which works if you plug in random values for x. "After" seeing the answer I reasoned maybe it could be broken down as (n+1)(n+2)n!/n!, but not really understanding why... like how am I factoring out the n! & ending up with that?

Here is another one I'm struggling with n!/(n-1)!. Here again, the book shows the answer to be n, which works, but can't quiet figure out how to get started. I thought maybe it'd be n!/(n-1)n! but that doesn't give me n as an answer.

 

[malawi_glenn]

well look at this:

4! = 4*3*2*1

n! = n*(n-1)*(n-2)* ... *2*1

So now, what do you think of the:
(n+2)!

 

[Feldoh]

I think it helps as malawi_glenn said to expand the factorials.

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

 

[kuahji]

Ok, that helps I think. So with say this one n!/(n-1)! that is like saying

n(n-1)(n-2).../(n-1)(n-2)... = n

Its like the denominator starts at (n-1) instead of n in the sequence & the rest just cancels out, correct? Thanks for the help.

 

[malawi_glenn]

Ok, that helps I think. So with say this one n!/(n-1)! that is like saying

n(n-1)(n-2).../(n-1)(n-2)... = n

Its like the denominator starts at (n-1) instead of n in the sequence & the rest just cancels out, correct? Thanks for the help.

Well, yes.
If you are not comfortable with doing it with n, try to do it with a number, say 5, in the begining, then you will get more comfortable with the more abstract ones.