Question -
Answer -
Given:
P(n, 5) = 20 P(n, 3)
By using the formula,
P (n, r) = n!/(n – r)!
P (n, 5) = n!/(n – 5)!
P (n, 3) = n!/(n – 3)!
So, from the question,
P(n, 5) = 20 P(n, 3)
Substituting theobtained values in above expression we get,
n!/(n – 5)! = 20 ×n!/(n – 3)!
Upon evaluating,
n! (n – 3)! / n! (n –5)! = 20
[(n –3) (n – 3 – 1) (n – 3 – 2)!] / (n – 5)! = 20
[(n –3) (n – 4) (n – 5)!] / (n – 5)! = 20
(n – 3) (n – 4) = 20
n2 –3n – 4n + 12 = 20
n2 –7n + 12 – 20 = 0
n2 –7n – 8 = 0
n2 –8n + n – 8 = 0
n(n – 8) – 1(n – 8) =0
(n – 8) (n – 1) = 0
n = 8 or 1
For, P(n, r): n ≥ r
∴ n =8 [for, P(n, 5)]