1. Chapter 7 Class 11 Permutations and Combinations
2. Serial order wise
3. Examples

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Example 12 Find the value of n such that (ii) "nP4" /"n 1P4" = 5/3 , n > 4 Lets first calculate nP4 and n 1P4 nP4 = !/( 4)! = ( ( 1)( 2)( 3)( 4)!)/( 4)! = n(n 1)(n 2)(n 3) Now "nP4" /"n 1P4" = 5/3 3 nP4 = 5 n-1P4 3(n)(n 1)(n 2)(n 3) = 5(n 1) (n 2) (n 3) (n 4) (3 ( 1)( 2)( 3))/(( 1)( 2)( 3)) = 5(n 4) 3n = 5(n 4) 3n = 5n 20 20 = 5n 3n 20 = 2n 20/2 = n 10 = n Hence n = 10. Example 12 Find the value of n such that nP5 = 42 nP3, n > 4 nP5 = 42 nP3 Calculating nP5 nP5 = !/( 5)! = ( ( 1)( 2)( 3)( 4)( 5)!)/( 5)! = n(n 1)(n 2)(n 3)(n 4) Now, it is given that nP5 = 42 nP3 n(n 1)(n 2)(n 3)(n 4) = 42n(n 1)(n 2) ( ( 1)( 2)( 3)( 4)( 5)!)/( ( 1)( 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 Hence, n = 10 & n = 3 But, It is given in question n > 4 So n = -3 not possible Therefore, n = 10 only

Examples