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Exercise 6.1 · Q9

Q.Find nn if n!3!(n−3)!:n!5!(n−5)!=5:3\dfrac{n!}{3!(n-3)!} : \dfrac{n!}{5!(n-5)!} = 5 : 3

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Solve the ratio equation n!3!(n−3)!:n!5!(n−5)!=5:3\dfrac{n!}{3!(n-3)!} : \dfrac{n!}{5!(n-5)!} = 5:3 — recognising both sides as combination formulas nC3^{n}C_3 and nC5^{n}C_5 — for the positive integer nn.

Combination formula: nCr=n!r!(n−r)!^{n}C_r = \dfrac{n!}{r!(n-r)!}. The given ratio is nC3:nC5^{n}C_3 : {}^{n}C_5. Their quotient simplifies using n!n!=1\dfrac{n!}{n!}=1 cancellation and factorial expansion: (n−3)!=(n−3)(n−4)(n−5)!(n-3)! = (n-3)(n-4)(n-5)!.

  1. Write the ratio as: nC3nC5=n!/[3!(n−3)!]n!/[5!(n−5)!]=n!3!(n−3)!×5!(n−5)!n!\dfrac{{}^{n}C_3}{{}^{n}C_5} = \dfrac{n!/[3!(n-3)!]}{n!/[5!(n-5)!]} = \dfrac{n!}{3!(n-3)!}\times\dfrac{5!(n-5)!}{n!}.
  2. Cancel n!n! from numerator and denominator: =5!(n−5)!3!(n−3)!= \dfrac{5!(n-5)!}{3!(n-3)!}.
  3. Compute 5!3!=1206=20\dfrac{5!}{3!} = \dfrac{120}{6} = 20.
  4. Expand (n−3)!=(n−3)(n−4)(n−5)!(n-3)! = (n-3)(n-4)(n-5)!, so (n−5)!(n−3)!=(n−5)!(n−3)(n−4)(n−5)!=1(n−3)(n−4)\dfrac{(n-5)!}{(n-3)!} = \dfrac{(n-5)!}{(n-3)(n-4)(n-5)!} = \dfrac{1}{(n-3)(n-4)}.
  5. Combine: nC3nC5=20(n−3)(n−4)\dfrac{{}^{n}C_3}{{}^{n}C_5} = \dfrac{20}{(n-3)(n-4)}.
  6. Set this equal to the given ratio 5:35:3, i.e. 53\dfrac{5}{3}: 20(n−3)(n−4)=53\dfrac{20}{(n-3)(n-4)} = \dfrac{5}{3}. …

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