Given:
5 P(4, n) = 6 P(5, n – 1)
By using the formula,
P (n, r) =\( \dfrac{ n!}{r!(n – r)!}\)
P (4, n) = \( \dfrac{ 4!}{r!(4 – r)!}\)
P (5, n-1) =\( \dfrac{5!}{(5 – (n-1))!}\)
= \( \dfrac{5!}{(5 – (n-1))!}\)
=\( \dfrac{5!}{(6 – n)!}\)
So, from the question,
5 P(4, n) = 6 P(5, n – 1)
Substituting the obtained values in above expression we get,
\(\dfrac{ 5 × 4!}{(4 – n)!} = \dfrac{6 × 5!}{(6 – n)!}\)
Upon evaluating,
\(\dfrac{ (6 – n) (5 – n) (4 – n)! }{ (4 – n)!} = 6 \)
(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)]
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