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Example 12 - Find value of n nP4 / n-1P4 = 5/3 and  nP5 = 42 nP3 - Examples

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  1. Chapter 7 Class 11 Permutations and Combinations
  2. Serial order wise
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Example 12 Find the value of n such that (ii) "nP4" /"nโˆ’1P4" = 5/3 , n > 4 Lets first calculate nP4 and n โ€“ 1P4 nP4 = ๐‘›!/(๐‘› โˆ’ 4)! = (๐‘›(๐‘›โˆ’1)(๐‘›โˆ’2)(๐‘›โˆ’3)(๐‘›โˆ’4)!)/(๐‘›โˆ’4)! = n(n โ€“ 1)(n โ€“ 2)(n โ€“ 3) Now "nP4" /"nโˆ’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‘› โˆ’1)(๐‘› โˆ’2)(๐‘› โˆ’ 3))/((๐‘› โˆ’1)(๐‘› โˆ’2)(๐‘› โˆ’ 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‘›โˆ’5)! = (๐‘›(๐‘›โˆ’1)(๐‘›โˆ’2)(๐‘›โˆ’3)(๐‘›โˆ’4)(๐‘›โˆ’5)!)/(๐‘›โˆ’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‘›โˆ’1)(๐‘›โˆ’2)(๐‘›โˆ’3)(๐‘›โˆ’4)(๐‘›โˆ’5)!)/(๐‘›(๐‘› โˆ’ 1)(๐‘› โˆ’ 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

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