n=4
P( 2n+1, n-1): P(2n-1,n) = 3:5
or, (2n+1)!/(n+2)!: (2n-1)!/(n-1)!=3/5 (using the permutations formula npr=n!/(n-r)!
or(2n+1)!/(n+2)! * (n-1)!/(2n-1)!=3/5
2n(2n+1)/n(n+1)(n+2)=3/5
solve this..you will get n=4. reject he other value i.e n=-1/3
If P( 2n+1, n-1): P(2n-1,n) = 3:5, find n
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tohellandback
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