Examples

Chapter 6 Class 11 Permutations and Combinations
Serial order wise

### Transcript

Example 12 Find the value of n such that (ii) "nP4" /"nā1P4" = 5/3 , n > 4 Lets first calculate nP4 and n ā 1P4 nP4 = š!/(š ā 4)! = (š(š ā 1)(š ā 2)(š ā 3)(š ā 4)!)/(š ā 4)! = n(n ā 1)(n ā 2)(n ā 3) n ā 1P4 = ((š ā 1)!)/(š ā 1 ā 4)! = ((š ā 1)!)/(š ā 5)! = ((š ā 1)(š ā 2)(š ā 3)(š ā 4)(š ā 5)!)/(š ā 5)! = (n ā 1) (n ā 2) (n ā 3) (n ā 4) nPr = ((š)!)/(š ā š)! Now "nP4" /"nā1P4" = 5/3 3 nP4 = 5 n-1P4 3(n)(n ā 1)(n ā 2)(n ā 3) = 5(n ā 1) (n ā 2) (n ā 3) (n ā 4) (3š(š ā1)(š ā2)(š ā 3))/((š ā1)(š ā2)(š ā 3)) = 5(n ā 4) 3n = 5(n ā 4) 3n = 5n ā 20 20 = 5n ā 3n 20 = 2n 20/2 = n 10 = n Hence, n = 10 n(n ā 10) + 3(n ā 10) = 0 (n ā 10) (n + 3) = 0 So, n = 10, and n = ā3 But, It is given in question n > 4 So n = ā3 not possible Therefore, n = 10 only Example 12 Find the value of n such that (ii) "nP4" /"nā1P4" = 5/3 , n > 4 Lets first calculate nP4 and n ā 1P4 nP4 = š!/(š ā 4)! = (š(š ā 1)(š ā 2)(š ā 3)(š ā 4)!)/(š ā 4)! = n(n ā 1)(n ā 2)(n ā 3) n ā 1P4 = ((š ā 1)!)/(š ā 1 ā 4)! = ((š ā 1)!)/(š ā 5)! = ((š ā 1)(š ā 2)(š ā 3)(š ā 4)(š ā 5)!)/(š ā 5)! = (n ā 1) (n ā 2) (n ā 3) (n ā 4) nPr = ((š)!)/(š ā š)! Now "nP4" /"nā1P4" = 5/3 3 nP4 = 5 n-1P4 3(n)(n ā 1)(n ā 2)(n ā 3) = 5(n ā 1) (n ā 2) (n ā 3) (n ā 4) (3š(š ā1)(š ā2)(š ā 3))/((š ā1)(š ā2)(š ā 3)) = 5(n ā 4) 3n = 5(n ā 4) 3n = 5n ā 20 20 = 5n ā 3n 20 = 2n 20/2 = n 10 = n Hence, n = 10