[SOLVED] If P( 2n+1, n-1): P(2n-1,n) = 3:5, find n

Status
Not open for further replies.
I

Ishaan Karnik

Member level 3
Joined
Jun 16, 2012
Messages
64
Helped
14
Reputation
28
Reaction score
14
Trophy points
1,288
Activity points
1,726
If P( 2n+1, n-1): P(2n-1,n) = 3:5, find n
Answer is 4 but how???
 

Let's start from P(a,b)=a!/(a-b)! that is:

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

then:

P(2n+1,n-1)/P(2n-1,n) = [(2n+1)!/(n+2)!]/[(2n-1)!/(n-1)!] = (2n+1)!*(n-1)!/[(2n-1)!*(n+2)!]

but:

(2n+1)! = (2n+1)*2n*(2n-1)!
(n+2)! = (n+2)*(n+1)*n*(n-1)!

thus:

P(2n+1,n-1)/P(2n-1,n) = (2n+1)*2n*(2n-1)!*(n-1)!/[(n+2)*(n+1)*n*(n-1)!*(2n-1)!]

simplifying:

P(2n+1,n-1)/P(2n-1,n) = (2n+1)*2/[(n+2)*(n+1)]

then:

(4n+2)/(n²+3n+2) = 3/5

20n+10=3n²+9n+6

3n²-11n-4=0

the acceptable solution is: (11+13)/6=4
the other one is negative.
 
Status
Not open for further replies.

Similar threads

Menu
Log in
Register
Cookies are required to use this site. You must accept them to continue using the site. Learn more…