Evaluating Fractions with fractorials

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The discussion focuses on evaluating factorial expressions, specifically (n+2)!/n! and n!/(n-1)!. The first expression simplifies to (n+2)(n+1) by canceling out n! from the numerator and denominator. The second expression simplifies to n, as the factorials cancel out all terms except for n. Participants suggest using specific numbers to understand the concepts better before tackling abstract forms. Overall, breaking down factorials into their components aids in understanding their simplification.
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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.
 
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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)!
 
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}
 
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.
 
kuahji said:
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.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
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