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Worked Examples · Example 6

Q.Convert the following products into factorial notation

(i) 2.4.6.8...(2n)2.4.6.8...(2n)
(ii) 1.3.5...(2n−1)1.3.5...(2n-1)
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Factor a 22 out of every even term for (i); split (2n)!(2n)! into its odd and even factors for (ii).

(2n)!=[1⋅2⋅3⋯(2n)]=[1⋅3⋅5⋯(2n−1)]×[2⋅4⋅6⋯(2n)](2n)! = \big[1\cdot2\cdot3\cdots(2n)\big] = \big[1\cdot3\cdot5\cdots(2n-1)\big]\times\big[2\cdot4\cdot6\cdots(2n)\big], splitting all integers up to 2n2n into odd and even ones.

  1. (i) 2⋅4⋅6⋯(2n)=(2×1)(2×2)(2×3)⋯(2×n)2\cdot4\cdot6\cdots(2n) = (2\times1)(2\times2)(2\times3)\cdots(2\times n), one factor of 22 from each of the nn terms.
  2. Pulling out the nn twos: =2n×(1×2×3×⋯×n)=2n n!= 2^n\times(1\times2\times3\times\cdots\times n) = 2^n\,n!.
  3. (ii) From the FORMULA band, [1⋅3⋅5⋯(2n−1)]=(2n)!2⋅4⋅6⋯(2n)\big[1\cdot3\cdot5\cdots(2n-1)\big] = \dfrac{(2n)!}{2\cdot4\cdot6\cdots(2n)}. …

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