Why does ((10-n)!)/((8-n)!) = (10-n)(9-n)?

[Erzeon]

How does ((10-n)!)/((8-n)!) = (10-n)(9-n)? I take my 2nd exam tomorrow and I came across this question

 

[quantumdude]

First state the definition of the factorial function, then apply it to (10-n)! .

 

[Erzeon]

What do you mean by definition?

Do you mean the whole entire equation?

If so, it was ((10-n)!)/((8-n)!) >= 9

then it somehow became (10-n)(9-n) >= 9

 

[Erzeon]

I think I found out, is it because if you expand the factorials out all the way to 0, the (8-n),(7-n),(6-n),(5-n),(4-n),(3-n),(2-n),(1-n) all cancel out in the numerator and denominator? So (10-n) and (9-n) is left?

 

[six789]

How does ((10-n)!)/((8-n)!) = (10-n)(9-n)? I take my 2nd exam tomorrow and I came across this question

that question is so easy!
you know the definition of factorial right? (n(n-1)...3*2*1)
so if you encounter that kind of problem just apply what you learn from the factorial...
Left Side

((10-n)!)-->> you can right this as ((10-n)!)((9-n)!)((8-n)!)...((1-n)!)
and ((8-n)!)-->> you can right this as ((8-n)!)...((1-n)!)

so, if you see ((8-n)!) in the numerator and denominator, it cancels out.
therefore left side = right side...
(10-n)(9-n) = (10-n)(9-n)

 

[Erzeon]

Thanks, I was blind to not see it.

 

[six789]

I think I found out, is it because if you expand the factorials out all the way to 0, the (8-n),(7-n),(6-n),(5-n),(4-n),(3-n),(2-n),(1-n) all cancel out in the numerator and denominator? So (10-n) and (9-n) is left?

exactly! see it is so easy right?

 

[six789]

Thanks, I was blind to not see it.

no problem... we are all here to benefit frm each others knowledge...

 

[quantumdude]

What do you mean by definition?

I meant just what I said: What is the definition of the factorial function function f(n)=n! .

no problem... we are all here to benefit frm each others knowledge...

Yes, but we do have rules here, which you all agreed to. We don't give assistance until the person asking the question shows an attempt at the problem. Guiding quesitons are OK, but complete solutions are not.

 

[six789]

Yes, but we do have rules here, which you all agreed to. We don't give assistance until the person asking the question shows an attempt at the problem. Guiding questions are OK, but complete solutions are not.

ok then mr. tom mattson...

 

[Erzeon]

Dont worry, I had an attempt at the solution before, I asked because I got stuck and didn't think to expand it. What I really needed was a quick answer because my final high school maths exam is on tomorrow and it counts towards my score that determines what courses I can get into.

Yeh it was easy lol, I tried to think of the answer last night while going to sleep but was too tired.:D

 

[six789]

lol... can u check my post, see if you can do it?

 

[Erzeon]

yeh I can, thanks to both of you:D

 

[six789]

can u check if it is correct.., reply on my post, not here...