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Proving factorial equations

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  1. Oct 16, 2015 #1
    • Member warned about posting without the template and work shown only in an image
    number 15 questions b and c are giving me a very hard time. I have tried expanding them then factoring out the common terms but somehow not getting it to be proven. detailed help will be appreciated.
     

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  3. Oct 16, 2015 #2

    RUber

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    I don't think factoring is the right answer.
    Are you sure that (b) is written correctly? n choose k is equal to n choose (n-k) normally, so I don't see how the sum could be.

    For (c), try writing it out and rearranging.

    ##\frac{n!}{k!(n-k)!} + \frac{n!}{(k-1)!(n-(k-1))!}##
    And you want this to be equal to:
    ##\frac{n+1!}{k!((n+1)-k)!}##

    **edit** You should try to make a common denominator to add the fractions. If you do this carefully and correctly, the right answer pops right out.
     
  4. Oct 16, 2015 #3
    yes b) is written correctly
     
  5. Oct 16, 2015 #4

    RUber

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    Just to be sure,
    (b) says: Show that n choose k + n choose (k-1) = n choose (n-k) ?
    That is straight wrong.
    n choose k = n choose (n-k). This is clear from the formula factorial representation.
    And n choose k + n choose (k-1) = (n+1) choose k, as in (c).
     
  6. Oct 17, 2015 #5

    Ray Vickson

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    I haven't read your question (since I don't read attachments), but I assume from what RUber has written that you want to show ##{}_nC_k + {}_nC_{k-1} = {}_nC_{n-k}##. As RUber has indicated, that is wrong. Try it for yourself: set n = 4, k = 2 and see what happens.

    The several-century-old Passcal triangle formula says that ##{}_n C _k + {n}C_{k-1} = {}_{n+1} C _k##. Again, post #2# shows you one way you can do it; another way is to start from ##(1+x)^n = \sum_{k=0}^n {}_nC_k x^k##, then write ##(1+x)^{n+1}## in two ways: one way is to use the previous expansion with ##n+1## in place of ##n##, and the other is to write it as ##(1+x) (1+x)^n##, use the expansion for the second factor, then gather together powers of ##x##.
     
  7. Oct 18, 2015 #6

    haruspex

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    From the text before and after b) in the attachment, it is very clear to me that b) ought to read "show that nCk=nCn-k."
     
  8. Oct 19, 2015 #7

    RUber

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    I wonder if @tesha is still working on this.
    If so, then I am sure the correction haruspex suggests will help. That interpretation for (b) will also be reinforced by (d), which looks to be a direct application of that rule.
     
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