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,
\(\dfrac{ P (n, 5) }{ P(n, 3)} = \dfrac{2 }{ 1}\)
Substituting the obtained values in above expression we get,
\( \dfrac{(n – 3) (n – 4) (n – 5)!}{ (n – 5)!} = \dfrac{2}{1}\)
(n – 3)(n – 4) = 2
n2 – 3n – 4n + 12 = 2
n2 – 7n + 12 – 2 = 0
n2 – 7n + 10 = 0
n2 – 5n – 2n + 10 = 0
n (n – 5) – 2(n – 5) = 0
(n – 5) (n – 2) = 0
n = 5 or 2
For, P (n, r): n ≥ r
n = 5 [for, P (n, 5)]
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