Given as

P(2n – 1, n) : P(2n + 1, n – 1) = 22 : 7

P(2n – 1, n) / P(2n + 1, n – 1) = 22/7

On using the formula,

P (n, r) = n!/(n – r)!

P (2n – 1, n) = (2n – 1)! / (2n – 1 – n)!

= (2n – 1)! / (n – 1)!

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

= (2n + 1)! / (n + 2)!

Therefore, from the question,

P(2n – 1, n) / P(2n + 1, n – 1) = 22/7

On substituting the obtained values in above expression we get,

[(2n – 1)! / (n – 1)!] / [(2n + 1)! / (n + 2)!] = 22/7

[(2n – 1)! / (n – 1)!] × [(n + 2)! / (2n + 1)!] = 22/7

[(2n – 1)! / (n – 1)!] × [(n + 2) (n + 2 – 1) (n + 2 – 2) (n + 2 – 3)!] / [(2n + 1) (2n + 1 – 1) (2n + 1 – 2)] = 22/7

[(2n – 1)! / (n – 1)!] × [(n + 2) (n + 1) n(n – 1)!] / [(2n + 1) 2n (2n – 1)!] = 22/7

[(n + 2) (n + 1)] / (2n + 1)2 = 22/7

7(n + 2) (n + 1) = 22×2 (2n + 1)

7(n² + n + 2n + 2) = 88n + 44

7(n² + 3n + 2) = 88n + 44

7n² + 21n + 14 = 88n + 44

7n² + 21n – 88n + 14 – 44 = 0

7n² – 67n – 30 = 0

7n² – 70n + 3n – 30 = 0

7n(n – 10) + 3(n – 10) = 0

(n – 10) (7n + 3) = 0

n = 10, -3/7
As we know that, n ≠ -3/7

∴ The value of n is 10.