anonymous
  • anonymous
MEDAL&FAN&Testimonials (n-r+1)!/(n-r-2)! How do I write this without using factorial notation?
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
My head's stuck on this problem if anyone could offer me help that would be awesome!
anonymous
  • anonymous
@ganeshie8 the almighty?
anonymous
  • anonymous
Hi

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anonymous
  • anonymous
Do you know how to do this problem?
ganeshie8
  • ganeshie8
Hint : let \(n-r+1 = m\)
anonymous
  • anonymous
The trickiest part is that denominator is negative whereas the numerator is positive.
anonymous
  • anonymous
This makes me feel insecure about canceling either.
anonymous
  • anonymous
(m)(m-1)(m-2)/(m-3)(m-4)(m-5)
ganeshie8
  • ganeshie8
\[\dfrac{(n-r+1)!}{(n-r-2)!} = \dfrac{(\color{blue}{n-r+1})!}{(\color{blue}{n-r+1}-3)!} \]
anonymous
  • anonymous
Oh!
anonymous
  • anonymous
Looks like the factorials are all canceled but (-3)!, at least intuitively.
ganeshie8
  • ganeshie8
Let that blue part equal \(\color{blue}{m}\) : \[\dfrac{(n-r+1)!}{(n-r-2)!} = \dfrac{(\color{blue}{n-r+1})!}{(\color{blue}{n-r+1}-3)!} =\dfrac{\color{blue}{m}!}{(\color{blue}{m}-3)!} =mP3\]
anonymous
  • anonymous
1/-3!
anonymous
  • anonymous
no no
anonymous
  • anonymous
How come mP3 and not 1/-3!?
ganeshie8
  • ganeshie8
recall the definition of permutation : \[nPr = \dfrac{n!}{(n-r)!}\]
anonymous
  • anonymous
Oh yes.
anonymous
  • anonymous
Not the conventional calculation of fractions.
anonymous
  • anonymous
Assuming that rule, (n+4)!/(n+2)! will be not (n+4)(n+3)?
anonymous
  • anonymous
(n+4)(n+3)(n+2)(n+1)(n)/(n+2)(n+1)(n)=(n+4)(n+3)? This is not correct?
UsukiDoll
  • UsukiDoll
\[\frac{(n+4)(n+3)(n+2)(n+1)(n)}{(n+2)(n+1)(n)}\] cancel like terms should be (n+4)(n+3)
anonymous
  • anonymous
At first seemed a bit counterintuitive after what Ganashie showed me.
anonymous
  • anonymous
But I guess I am not so retarded at this timeXD

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