Given as
5 P(4, n) = 6 P(5, n – 1)
On using the formula,
P (n, r) = n!/(n – r)!
P (4, n) = 4!/(4 – n)!
P (5, n-1) = 5!/(5 – (n-1))!
= 5!/(5 – n + 1)!
= 5!/(6 – n)!
Therefore, from the question,
5 P(4, n) = 6 P(5, n – 1)
On substituting the obtained values in above expression we get,
5 × 4!/(4 – n)! = 6 × 5!/(6 – n)!
Now, upon evaluating,
(6 – n)! / (4 – n)! = 6/5 × 5!/4!
[(6 – n) (6 – n – 1) (6 – n – 2)!] / (4 – n)! = (6 × 5 × 4!) / (5 × 4!)
[(6 – n) (5 – n) (4 – n)!] / (4 – n)! = 6
(6 – n) (5 – n) = 6
30 – 6n – 5n + n2 = 6
30 – 6 – 11n + n2 = 0
n2 – 11n + 24 = 0
n2 – 8n – 3n + 24 = 0
n(n – 8) – 3(n – 8) = 0
(n – 8) (n – 3) = 0
n = 8 or 3
For, P (n, r): r ≤ n
∴ n = 3 [for, P (4, n)]
