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Exercise 6.1 · Q7

Q.Convert into factorials

(i) 5.6.7.85.6.7.8
(ii) 3.6.9.12.153.6.9.12.15
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Express two given numerical products as compact factorial expressions — a run of consecutive integers as a ratio of factorials, and a run of multiples of 3 as 353^5 times a smaller factorial.

Two useful factorial identities:

  • A product of consecutive integers from aa to bb (i.e. a(a+1)(a+2)⋯ba(a+1)(a+2)\cdots b) equals b!(a−1)!\dfrac{b!}{(a-1)!}, since b!=b(b−1)⋯a⋅(a−1)!b! = b(b-1)\cdots a\cdot(a-1)!.
  • A product of equally-spaced multiples of dd, i.e. d⋅(2d)⋅(3d)⋯(kd)d\cdot(2d)\cdot(3d)\cdots(kd), equals dk×(1⋅2⋅3⋯k)=dk k!d^k\times(1\cdot2\cdot3\cdots k) = d^k\,k!, by factoring dd out of each term.

(i) Convert 5⋅6⋅7⋅85\cdot6\cdot7\cdot8

  1. This is a product of 4 consecutive integers running from a=5a=5 to b=8b=8.
  2. By the consecutive-run identity with a=5,b=8a=5,b=8: 5⋅6⋅7⋅8=8!(5−1)!=8!4!5\cdot6\cdot7\cdot8 = \dfrac{8!}{(5-1)!} = \dfrac{8!}{4!}.
  3. Verify numerically: 8!=403208!=40320, 4!=244!=24, so 8!4!=4032024=1680\dfrac{8!}{4!} = \dfrac{40320}{24} = 1680. Direct multiplication: 5×6=305\times6=30, 30×7=21030\times7=210, 210×8=1680210\times8=1680. ✓ Matches.

(ii) Convert 3⋅6⋅9⋅12⋅153\cdot6\cdot9\cdot12\cdot15 …

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