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

Transcript

Example 12 Find the value of n such that (ii) "nP4" /"nโ1P4" = 5/3 , n > 4 Lets first calculate nP4 and n โ 1P4 nP4 = ๐!/(๐ โ 4)! = (๐(๐โ1)(๐โ2)(๐โ3)(๐โ4)!)/(๐โ4)! = n(n โ 1)(n โ 2)(n โ 3) Now "nP4" /"nโ1P4" = 5/3 3 nP4 = 5 n-1P4 3(n)(n โ 1)(n โ 2)(n โ 3) = 5(n โ 1) (n โ 2) (n โ 3) (n โ 4) (3๐(๐ โ1)(๐ โ2)(๐ โ 3))/((๐ โ1)(๐ โ2)(๐ โ 3)) = 5(n โ 4) 3n = 5(n โ 4) 3n = 5n โ 20 20 = 5n โ 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 = ๐!/(๐โ5)! = (๐(๐โ1)(๐โ2)(๐โ3)(๐โ4)(๐โ5)!)/(๐โ5)! = n(n โ 1)(n โ 2)(n โ 3)(n โ 4) Now, it is given that nP5 = 42 nP3 n(n โ 1)(n โ 2)(n โ 3)(n โ 4) = 42n(n โ 1)(n โ 2) (๐(๐โ1)(๐โ2)(๐โ3)(๐โ4)(๐โ5)!)/(๐(๐ โ 1)(๐ โ 2)!) = 42 (n โ 3)(n โ 4) = 42 n(n - 4) โ 3(n โ 4) = 42 n2 โ 4n โ 3n + 12 = 42 n2 โ 7n + 12 = 42 n2 โ 7n + 12 โ 42 = 0 n2 โ 10n + 3n โ 30 = 0 n(n โ 10) + 3(n โ 10) = 0 (n โ 10) (n + 3) = 0 Hence, n = 10 & n = โ 3 But, It is given in question n > 4 So n = -3 not possible Therefore, n = 10 only