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

  • Thread starter six789
  • Start date
  • #1
127
0
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
 

Answers and Replies

  • #2
185
4
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.
 
  • #3
127
0
permutations

qbert said:
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
 
Last edited:
  • #4
127
0


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 dont know wat do do next, im soo stuck, since you cannot cancel any of them, so should i cross multiply it?
 

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

  • Last Post
Replies
16
Views
3K
  • Last Post
Replies
22
Views
9K
Replies
3
Views
5K
Replies
4
Views
3K
  • Last Post
2
Replies
26
Views
3K
Replies
5
Views
8K
Replies
3
Views
2K
Replies
6
Views
3K
Top