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Ex 7.4, 2 - Determine n if (i) 2nC3 : nC3 = 12 : 1 - Chapter 7 - Combination formula

  1. Chapter 7 Class 11 Permutations and Combinations
  2. Serial order wise
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Ex 7.4, 2 Determine n if 2nC3 : nC3 = 12 : 1 Let first calculate 2nC3 and nC3 separately 2nC3 = 2๐‘›!/3!(2๐‘›โˆ’3)! = ((2๐‘›)(2๐‘› โˆ’1)(2๐‘›โˆ’2)(2๐‘›โˆ’3)!)/((3 ร— 2 ร— 1)(2๐‘›โˆ’3)!) = 2๐‘›(2๐‘›โˆ’1)(2๐‘›โˆ’2)/6 Given 2nC3 : nC3 = 12 : 1 2๐‘›(2๐‘›โˆ’1)(2๐‘›โˆ’2)/6 : ๐‘›(๐‘›โˆ’1)(๐‘›โˆ’2)/6 = 12 : 1 2๐‘›(2๐‘›โˆ’1)(2๐‘›โˆ’2)/6 : ๐‘›(๐‘›โˆ’1)(๐‘›โˆ’2)/6 = 12 : 1 ((2๐‘›(2๐‘› โˆ’1)(2๐‘› โˆ’2))/6)/(๐‘›(๐‘›โˆ’1)(๐‘›โˆ’2)/6) = 12 2n(2nโˆ’1)(2nโˆ’2)/6 ร— 6/(๐‘›(๐‘› โˆ’ 1)(๐‘› โˆ’ 2)) = 12 2n(2nโˆ’1)(2nโˆ’2)/(๐‘›(๐‘› โˆ’ 1)(๐‘› โˆ’ 2)) ร— 6/6 = 12 (2(2n โˆ’ 1). 2(n โˆ’ 1))/((๐‘› โˆ’ 1)(๐‘› โˆ’ 2)) = 12 4(2n โˆ’ 1)(n โˆ’ 1)/((๐‘› โˆ’ 1)(๐‘› โˆ’ 2)) = 12 4(2n โˆ’ 1)/((๐‘› โˆ’ 2)) = 12 4(2n โ€“ 1) = 12(n โ€“ 2) 8n โ€“ 4 = 12n โ€“ 24 โ€“ 4 + 24 = 12n โ€“ 8n 20 = 4n 20/4 = n 5 = n Thus, n = 5 Ex 7.4, 2 Determine n if (ii) 2nC3 : nC3 = 11 : 1 Let first calculate 2nC3 and nC3 separately 2nC3 = 2๐‘›!/3!(2๐‘›โˆ’3)! = ((2๐‘›)(2๐‘› โˆ’1)(2๐‘›โˆ’2)(2๐‘›โˆ’3)!)/((3 ร— 2 ร— 1)(2๐‘›โˆ’3)!) = 2๐‘›(2๐‘›โˆ’1)(2๐‘›โˆ’2)/6 Given 2nC3 : nc3 = 11 : 1 2๐‘›(2๐‘›โˆ’1)(2๐‘›โˆ’2)/6 : ๐‘›(๐‘›โˆ’1)(๐‘›โˆ’2)/6 = 11 : 1 2๐‘›(2๐‘›โˆ’1)(2๐‘›โˆ’2)/6 : ๐‘›(๐‘›โˆ’1)(๐‘›โˆ’2)/6 = 11 : 1 ((2๐‘›(2๐‘› โˆ’1)(2๐‘› โˆ’2))/6)/(๐‘›(๐‘›โˆ’1)(๐‘›โˆ’2)/6) = 11 2n(2nโˆ’1)(2nโˆ’2)/6 ร— 6/(๐‘›(๐‘› โˆ’ 1)(๐‘› โˆ’ 2)) = 11 2n(2nโˆ’1)(2nโˆ’2)/(๐‘›(๐‘› โˆ’ 1)(๐‘› โˆ’ 2)) ร— 6/6 = 11 (2(2n โˆ’ 1). 2(n โˆ’ 1))/((๐‘› โˆ’ 1)(๐‘› โˆ’ 2)) = 11 4(2n โˆ’ 1)(n โˆ’ 1)/((๐‘› โˆ’ 1)(๐‘› โˆ’ 2)) = 11 4(2n โˆ’ 1)/((๐‘› โˆ’ 2)) = 11 4(2n โ€“ 1) = 11(n โ€“ 2) 8n โ€“ 4 = 11n โ€“ 22 โ€“ 4 + 22 = 11n โ€“ 8n 18 = 3n 18/3 = n 6 = n Thus, n = 6

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