anonymous
  • anonymous
MEDAL&FAN&Testimonials (n-r+1)!/(n-r-2)! How do I write this without using factorial notation?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.