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

[Ishaan Karnik]

If P( 2n+1, n-1): P(2n-1,n) = 3:5, find n
Answer is 4 but how???

 

[albbg]

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.