## anonymous one year ago MEDAL&FAN&Testimonials (n-r+1)!/(n-r-2)! How do I write this without using factorial notation?

1. anonymous

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

2. anonymous

@ganeshie8 the almighty?

3. anonymous

Hi

4. anonymous

Do you know how to do this problem?

5. ganeshie8

Hint : let $$n-r+1 = m$$

6. anonymous

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

7. anonymous

This makes me feel insecure about canceling either.

8. anonymous

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

9. ganeshie8

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

10. anonymous

Oh!

11. anonymous

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

12. ganeshie8

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

13. anonymous

1/-3!

14. anonymous

no no

15. anonymous

How come mP3 and not 1/-3!?

16. ganeshie8

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

17. anonymous

Oh yes.

18. anonymous

Not the conventional calculation of fractions.

19. anonymous

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

20. anonymous

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

21. 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)

22. anonymous

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

23. anonymous

But I guess I am not so retarded at this timeXD