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Example 12 Find the value of n such that nP5 = 42 nP3, n > 4 Given nP5 = 42 nP3 Calculating nP5 nP5 = 𝑛!/(𝑛 − 5)! = (𝑛(𝑛 − 1)(𝑛 − 2)(𝑛 − 3)(𝑛 − 4)(𝑛 − 5)!)/(𝑛 − 5)! = n(n – 1)(n – 2)(n – 3)(n – 4) Calculating 42nP3 42nP3 = 42𝑛!/(𝑛 − 3)! = 42𝑛(𝑛 −1)(𝑛 − 2)(𝑛 − 3)!/(𝑛 − 3)! = 42n(n – 1)(n – 2) Now, nP5 = 42 nP3 n(n – 1)(n – 2)(n – 3)(n – 4) = 42n(n – 1)(n – 2) (𝑛(𝑛 − 1)(𝑛 − 2)(𝑛 − 3)(𝑛 − 4) )/(𝑛(𝑛 − 1)(𝑛 − 2) ) = 42 (n – 3)(n – 4) = 42 n(n – 4) – 3(n – 4) = 42 n2 – 4n – 3n + 12 = 42 n2 – 7n + 12 = 42 n2 – 7n + 12 – 42 = 0 n2 – 10n + 3n – 30 = 0 n(n – 10) + 3(n – 10) = 0 (n – 10) (n + 3) = 0 So, n = 10, and n = – 3 n(n – 10) + 3(n – 10) = 0 (n – 10) (n + 3) = 0 So, n = 10, and n = –3 But, It is given in question n > 4 So n = –3 not possible Therefore, n = 10 only

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.