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Worked Examples · Example 2

Q.Compute

(i) 11!9!\dfrac{11!}{9!}
(ii) 13!(2!)(11!)\dfrac{13!}{(2!)(11!)}
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✓ Free question

Cancel the common factorial in the denominator, then simplify the remaining product.

n!r!=n×(n−1)×⋯×(r+1)\dfrac{n!}{r!} = n\times(n-1)\times\cdots\times(r+1) for n≥rn \ge r, obtained by cancelling r!r! from n!=n×(n−1)×⋯×(r+1)×r!n! = n\times(n-1)\times\cdots\times(r+1)\times r!.

  1. (i) 11!9!=11×10×9!9!=11×10=110\dfrac{11!}{9!} = \dfrac{11\times10\times9!}{9!} = 11\times10 = 110.
  2. (ii) 13!2! 11!=13×12×11!2!×11!=13×122!=1562=78\dfrac{13!}{2!\,11!} = \dfrac{13\times12\times11!}{2!\times11!} = \dfrac{13\times12}{2!} = \dfrac{156}{2} = 78.
✓Final answer

(i) 11!9!=110\dfrac{11!}{9!} = 110 (ii) 13!2! 11!=78\dfrac{13!}{2!\,11!} = 78

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