# Solve the expression for n Є N

1. Nov 5, 2005

### six789

i just want to confirm if my answer is right....
this is the problem:

solve the expression for n Є N

P(2n + 4, 3) = 2/3P(n+4, 4)

this is my work:

(2n +4)!/((2n +4)-3)!) = (2/3(n+4)!)/((n+4)!)-4)!)
(2n +4)!/((2n +1)! )= (2/3(n+4)!)/(n!)

and i dont know wat to do next... should i just cross multiply it?

Last edited: Nov 5, 2005
2. Nov 5, 2005

### six789

permutations

help me with this if my answer is correct....
help is appreciated...

Last edited: Nov 5, 2005
3. Nov 6, 2005

### six789

help from anyone with this problem...

Last edited: Nov 6, 2005
4. Nov 6, 2005

### Gale

start canceling terms. if you have $$\frac{(2n+4)!}{(2n+1)!}$$ and you understand what factorial means, you can cancel all the factors in the denominator with some in the numerator. do this as well to the right side of your equation. then try to solve.

5. Nov 6, 2005

### six789

so you mean $$\frac{(2n+4)!}{(2n+1)!}$$ is same thing as writing as (2n+4)(2n+3)(2n+2)(2n+1)/(2n+1)?

6. Nov 6, 2005

### Gale

uh, well the (2n+1)'s cancel out. but ya. thats how a factorial works. so do the same to other side and go.

7. Nov 6, 2005

### six789

correct me if im wrong ok? i've try the other side... should you make the numerator 2/3(n+4)! to -2/3(n-4)! in order to cancel it from n(n-1)(n-2)(n-3)(n-4)?

8. Nov 6, 2005

### Gale

no... i think you're missing something...

$$\frac{2/3(n+4)!}{n!}= \frac{2/3(n+4)(n+3)(n+2)(n+1)(n)(n-1)(n-2)(n-3).....}{n(n-1)(n-2)(n-3)......}$$
so you're left with only $$2/3(n+4)(n+3)(n+2)(n+1)$$ cause you can cancel the rest out. do you see this?? does it make sense? this is what we did to the other side as well.

9. Nov 6, 2005

### six789

now i get gale... thanks so m uch, and it is clearer for me now...

10. Nov 6, 2005

### six789

what if you get like this.... -24(2n+3)(n+4)(n+3) = 0...
what should i do to the -24? or it is just 0, then the value for n is just n=-2/3, n=-4 and n=-3?

11. Nov 6, 2005

### Gale

well, i'm not really sure how you got that.... its not what i get. what's your equation after you've simplified? i get something simple enough to multiply out and then combine like terms. then solve.

12. Nov 6, 2005

### six789

i got like (2n+4)(2n+3)(2n+2) = 2/3(n+4)(n+3)(n+2)(n+1)
then... 2(n+2)(2n+3)2(n+1) = 2/3(n+4)(n+3)(n+2)(n+1)
which is 4(2n+3) = 2/3(n+4)(n+3)
and i dont know wat to do next

13. Nov 6, 2005

### Gale

multiply it out and simplify, its just a quadractic.