Question

MEDAL&FAN&Testimonials

(n-r+1)!/(n-r-2)!
How do I write this without using factorial notation?

Answer 1

My head's stuck on this problem if anyone could offer me help that would be awesome!

Answer 2

@ganeshie8 the almighty?

Answer 3

Hi

Answer 4

Do you know how to do this problem?

Answer 5

Hint :

let \(n-r+1 = m\)

Answer 6

The trickiest part is that denominator is negative whereas the numerator is positive.

Answer 7

This makes me feel insecure about canceling either.

Answer 8

(m)(m-1)(m-2)/(m-3)(m-4)(m-5)

Answer 9

\[\dfrac{(n-r+1)!}{(n-r-2)!} = \dfrac{(\color{blue}{n-r+1})!}{(\color{blue}{n-r+1}-3)!} \]

Answer 10

Oh!

Answer 11

Looks like the factorials are all canceled but (-3)!, at least intuitively.

Answer 12

Let that blue part equal \(\color{blue}{m}\) :
\[\dfrac{(n-r+1)!}{(n-r-2)!} = \dfrac{(\color{blue}{n-r+1})!}{(\color{blue}{n-r+1}-3)!} =\dfrac{\color{blue}{m}!}{(\color{blue}{m}-3)!} =mP3\]

Answer 13

1/-3!

Answer 14

no no

Answer 15

How come mP3 and not 1/-3!?

Answer 16

recall the definition of permutation :
\[nPr = \dfrac{n!}{(n-r)!}\]

Answer 17

Oh yes.

Answer 18

Not the conventional calculation of fractions.

Answer 19

Assuming that rule, (n+4)!/(n+2)! will be not (n+4)(n+3)?

Answer 20

(n+4)(n+3)(n+2)(n+1)(n)/(n+2)(n+1)(n)=(n+4)(n+3)? This is not correct?

Answer 21

\[\frac{(n+4)(n+3)(n+2)(n+1)(n)}{(n+2)(n+1)(n)}\] cancel like terms should be (n+4)(n+3)

Answer 22

At first seemed a bit counterintuitive after what Ganashie showed me.

Answer 23

But I guess I am not so retarded at this timeXD