Question -
Answer -
Given:
n+5Pn+1 = 11(n – 1)/2 n+3Pn
P (n +5, n + 1) = 11(n– 1)/2 P(n + 3, n)
By using the formula,
P (n, r) = n!/(n – r)!
(n + 5) (n + 4) = 22(n – 1)
n2 +4n + 5n + 20 = 22n – 22
n2 +9n + 20 – 22n + 22 = 0
n2 –13n + 42 = 0
n2 –6n – 7n + 42 = 0
n(n – 6) – 7(n – 6) =0
(n – 7) (n – 6) = 0
n = 7 or 6
∴ The value of ncan either be 6 or 7.