Example 12 (i) - Chapter 6 Class 11 Permutations and Combinations
Last updated at April 16, 2024 by Teachoo
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Example 6 (i)
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Example 12 (i) Important You are here
Example 12 (ii)
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Example 20 Important
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Chapter 6 Class 11 Permutations and Combinations
Serial order wise
- Ex 6.1
- Ex 6.2
- Ex 6.3
- Ex 6.4
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Transcript
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
