Given:
P(n, 5) = 20 P(n, 3)
By using the formula,
P (n, r) = \( \dfrac{ n!}{r!(n – r)!}\)
P (n, 5) = \(\dfrac{ n!}{r!(n – 5)!}\)
P (n, 3) = \( \dfrac{ n!}{r!(n – 3)!}\)
So, from the question,
P(n, 5) = 20 P(n, 3)
Substituting the obtained values in above expression we get,
\( \dfrac{ n!}{r!(n – 5)!}= 20 ×\dfrac{ n!}{(n – 3)!}\)
Upon evaluating,
\(\dfrac{ (n – 3) (n – 4) (n – 5)!}{ (n – 5)! }= 20\)
(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)]
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