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Algebraically Determining the Solution to this Combination

  1. May 22, 2015 #1
    1. The problem statement, all variables and given/known data

    [tex] \dbinom{7}{r} = 21 [/tex]

    2. Relevant equations

    [tex] \dbinom{n}{r} = \frac{n!}{(n-r)!r!} [/tex]

    3. The attempt at a solution

    [tex] \dbinom{7}{r} = 21 [/tex]

    [tex] \frac{7!}{(7-r)!r!} = 21 [/tex]

    [tex] 7! = 21(7-r)!r! [/tex]

    [tex] 240 = (7-r)!r! [/tex]

    So I get here and the convention is to guess in check (to my knowledge). I found a way to do it on my TI-84; simply go y= 7 nCr x ; then observe the table to gather the solutions of 5 and 2. But how would you do this algebraically (no guess and check)?
     
  2. jcsd
  3. May 22, 2015 #2

    ehild

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    Write out both the factorials and 21 in product form.
    ##7! =2\cdot3\cdot4\cdot5\cdot6\cdot7##.
    ##21=3\cdot7##.
    See what factors should cancel.
     
  4. May 22, 2015 #3
    What do you mean?
     
  5. May 22, 2015 #4

    ehild

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    Think of the definition of n!.
    [tex]\dbinom{7}{r} = \frac{2\cdot3\cdot4\cdot5\cdot6\cdot7}{(2\cdot3\cdot \cdot\cdot)(\cdot\cdot\cdot)}=3\cdot7[/tex]

    Simplifying the fraction, you get 7*3.
     
  6. May 22, 2015 #5
    Sorry, I still don't quite follow what's happening in the denominator.
     
  7. May 22, 2015 #6

    Mark44

    Staff: Mentor

    ehild is asking you to fill in the missing parts, using the definition of ##\dbinom{7}{r}##. What should be the last (largest) number in (##2 \cdot 3 \cdot \dots ?##)? What should be the factors in (##\dots##)?
     
  8. May 22, 2015 #7
    I think I have some clarity on it. In the numerator, the only pairs of numbers that even have a chance to be divided (once multiplied together) to make 21 are 7 and 6. So that means that 5! must cancel up in the numerator to leave these two factors. By consequence, (7-5)! = 2! = 2, thus 42/2 = 21. It's also then obvious that if r=2, 5! occurs in the (7-r)!, and also works. Does this reasoning make sense?
     
  9. May 23, 2015 #8

    ehild

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    Yes.
    Yes. But why do you multiply the factors? 21 = 3*7 and you have to divide 6*7 by 2 to get it.
     
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