Probability solve the expression P(2n+4,3) = 2/3P(n+4,4) for n

[six789]

i just want to confirm if my anser 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)!/(n-3)! = (2/3(n+4)!)/(n-4)!
-2(n-2)!/(n-3)! = -2/3(n-4)!/(n-4)!
-2/(n-3)! = -2/3
-2 = -2/3(n-3)
-2 = -2n/3 +6/3
-2 = -2n/3 +2
-6 = -2n +2
-8 = -2n
n=4

 

[qbert]

if by P(x,y) you mean # of permutations of x objects taken y at a time,
so the formula for P(x,y) = x!/(x-y)!
then no, your solution doesn't work.

P(2n + 4, 3) = 2/3P(n+4, 4)
this is my work:
(2n +4)!/(n-3)! = (2/3(n+4)!)/(n-4)!

P(2n+4,3) = (2n+4)!/((2n+4) - 3)! = (2n+4)!/(2n+1)!
same mistake on the other side of the equation.
plus, you should at least say what P(x,y) is, otherwise
no one knows what you're asking about.

 

[six789]

permutations

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

ok, i get what you mean, but how can i manipulate this, is seems i cannot cancel it... help again

 

[six789]

this wat i did...

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

i don't know wat do do next, I am soo stuck, since you cannot cancel any of them, so should i cross multiply it?