Given:

P (n, 4) = 12. P (n, 2)

By using the formula,

P (n, r) =\( \dfrac{ n!}{r!(n – r)!}\)

P (n, 4) = \( \dfrac{ n!}{r!(n – 4)!}\)

P (n, 2) =\( \dfrac{ n!}{r!(n – 2)!}\)

So, from the question,

P (n, 4) = 12. P (n, 2)

Substituting the obtained values in above expression we get,

\(\dfrac{ n!}{(n – 4)!} = 12 × \dfrac{n!}{(n – 2)!}\)

Upon evaluating,

(n – 2) (n – 3) = 12

n^{2} – 3n – 2n + 6 = 12

n^{2} – 5n + 6 – 12 = 0

n^{2} – 5n – 6 = 0

n^{2} – 6n + n – 6 = 0

n (n – 6) – 1(n – 6) = 0

(n – 6) (n – 1) = 0

n = 6 or 1

For, P (n, r): n ≥ r

n = 6 [for, P (n, 4)]

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