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Showing that (m n) + (m n-1) = (m+1 n).

  1. Apr 7, 2012 #1
    1. The problem statement, all variables and given/known data

    Show that if n is a positive integer at most equal to m, then
    (m n) + (m n-1) = (m+1 n).

    2. Relevant equations



    3. The attempt at a solution

    I understand that (m n) = m!/n!(m-n)!. I'm not entirely sure how to figure out (m n-1) because the book I'm studying from never explains this. It merely explains factorials and then shows that (m n) = m!/n!(m-n)!; however, I can at least figure out by reasoning it out that (m n-1) = m!/(m-n+1)!(n-1)!. However, I cannot understand how the next step goes.

    If we're adding m!/n!(m-n)! and m!/(m-n+1)!(n-1)! then I'm at a loss. The book says that the common denominator is n!(m-n+1)!, but I don't understand that. And the numerator comes out to be m!(m-n+1)+m!n?! This I just cannot understand! I've turned this problem over in my head for over and hour and still nothing! Please help!
     
  2. jcsd
  3. Apr 7, 2012 #2
    This problem is from Serge Lang's 'Basic Mathematics' book. Chapter 1 - Numbers >> Rational Numbers >> Problem 7, part d.
     
  4. Apr 7, 2012 #3

    rock.freak667

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    Since you have the two terms on the left expanded try expanding the individual terms.

    Your book should say that n!= n(n-1)(n-2)(n-3)...x3x2x1 right?

    Well this can be written in a simpler manner, n!=n(n-1)!

    See if you can simplify and bring both terms to a common denominator.
     
  5. Apr 7, 2012 #4
    The book doesn't say anything about how n! = n(n-1)!. But that confuses me. I'm new to factorials and the binomial coefficient, so I'm a little confounded at the moment. How is n! = n(n-1)! a simplification of n! = n(n-1)(n-2)(n-3)...x3x2x1? :O
     
  6. Apr 7, 2012 #5
    What I mean to say is how can you drop all of those other terms?
     
  7. Apr 7, 2012 #6
    Also, if I write n! as n(n-1), then I get: m!/n(n-1)(m-n)! + m!/(m-n+1)!(n-1)!. I still can't quite make both denominators fit... I'm sorry for bothering you so much over a simple problem like this... I promise I'm trying. I think I'm just cracking under pressure.
     
  8. Apr 7, 2012 #7
    Hmm... Even if I can get both denominators to become the same, how does the numerator end up as: m!(m-n+1)+m!n?! If they have a common denominator, shouldn't I be able to just add m! and m! together?
     
  9. Apr 7, 2012 #8

    Dick

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    If you don't understand why n!=n*(n-1)! then you are cracking under pressure. That's the most basic thing about factorials. 3*2*1=3*(2*1). So 3!=3*2!.
     
  10. Apr 7, 2012 #9
    Oooooh. I see! Sorry. XD I get how n! = n(n-1)!. Goshhhh. How silly.
     
  11. Apr 7, 2012 #10
    So then m! = m(m-1) as well. So I can also set m! = m(m-1) as both numerators. But there are still some steps I seem to be missing. Blargh, what am I missing?! ;___; And why is this only introduced as a problem with no explanation?
     
  12. Apr 7, 2012 #11
    I've been working through a bunch of different proofs and maybe I just need a break. I'm a bit discouraged that I don't understand what I'm doing when it comes to this part, but maybe it's alright. I just really want to be a theoretical physicist and it certainly is silly that such problems are slowing me down!
     
  13. Apr 7, 2012 #12

    Dick

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    Ok then the rest of it goes the same. Express (m, n-1) and (m+1, n) as some factor times (m,n). Then show you have an identity.
     
  14. Apr 7, 2012 #13
    I'm already nineteen and I'm still struggling with this stuff! Is it even possible for me to grasp differential geometry and tensor analysis in my lifetime at this rate?!
     
  15. Apr 7, 2012 #14
    I don't understand what you mean. Where are you getting (m, n-1) and (m+1, n)? ;__;
     
  16. Apr 7, 2012 #15

    Dick

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    Becoming a theoretical physicist involves a number of challenges and this isn't the greatest of them by a long shot. It's not alright if you can't do this. Do it.
     
  17. Apr 7, 2012 #16
    Oh, sorry. I meant it's alright if I take a break to cool my head. XD No, I won't leave any stones upright in this area!
     
  18. Apr 7, 2012 #17

    Dick

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    I'm getting them from your problem statement. Don't you remember?
     
  19. Apr 7, 2012 #18
    Oh, the commas confused me. The way it's written there were no commas. I see. Let me try re-starting the problem.
     
  20. Apr 7, 2012 #19
    (m, n) + (m, n-1) = (m+1, n). That's what I'm supposed to prove. So (m, n) = m!/n!(m-n!) and (m, n-1) = m!/(m-n+1)!(n-1)!. If I expand it, then I get: m(m-1)/n(n-1)(m-n)! + m(m-1)/(m-n+1)!(n-1)!. However, then I get stuck.
     
  21. Apr 7, 2012 #20
    I know that (m+1, n) = (m+1)!/n!((m+1)-n)!. But I can't get to this because I am still not sure what to do after I hit what was mentioned above...
     
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