(2n+2)Cn / 2nCn = 14/5

(2n+2)!/n!(2n+2-n)! * n!(2n-n)!/(2n)! = 14/5

(2n+2)!/n!(n+2)! * (n!)^2 / (2n)! = 14/5

(2n+2)(2n+1)/(n+2)(n+1) = 14/5

5(2n+2)(2n+1) = 14(n+2)(n+1)

5(4n^2 + 6n + 2) = 14(n^2 + 3n + 2)

20n^2 + 30n + 10 = 14n^2 + 42n + 28

6n^2 - 12n - 18 = 0

n^2 - 2n - 3 = 0

(n-3)(n+1) = 0

n = -1, 3 but n>0

n = 3

EDIT: It's meant to be perms not combs but still gives the same answer as dividing those 2 combinations gets rid of the n! which is the difference between the P and C.

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