If P(n, 5) : P(n, 3) = 2 : 1, find n.

1.	If P(n, 5) : P(n, 3) = 2 : 1, find n.
Answer» Given as P(n, 5) : P(n, 3) = 2 : 1 P(n, 5) / P(n, 3) = 2/1 On using the formula, P (n, r) = n!/(n – r)! P (n, 5) = n!/ (n – 5)! P (n, 3) = n!/ (n – 3)! Therefore, from the question, P (n, 5) / P(n, 3) = 2/1 On substituting the obtained values in above expression we get, [n!/ (n – 5)!] / [n!/ (n – 3)!] = 2/1 [n!/ (n – 5)!] × [(n – 3)! / n!] = 2/1 (n – 3)! / (n – 5)! = 2/1 [(n – 3) (n – 3 – 1) (n – 3 – 2)!] / (n – 5)! = 2/1 [(n – 3) (n – 4) (n – 5)!] / (n – 5)! = 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)]

Answer»

Given as

P(n, 5) : P(n, 3) = 2 : 1

P(n, 5) / P(n, 3) = 2/1

On using the formula,

P (n, r) = n!/(n – r)!

P (n, 5) = n!/ (n – 5)!

P (n, 3) = n!/ (n – 3)!

Therefore, from the question,

P (n, 5) / P(n, 3) = 2/1

On substituting the obtained values in above expression we get,

[n!/ (n – 5)!] / [n!/ (n – 3)!] = 2/1

[n!/ (n – 5)!] × [(n – 3)! / n!] = 2/1

(n – 3)! / (n – 5)! = 2/1

[(n – 3) (n – 3 – 1) (n – 3 – 2)!] / (n – 5)! = 2/1

[(n – 3) (n – 4) (n – 5)!] / (n – 5)! = 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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