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Solve the expression for n Є N

  1. Nov 5, 2005 #1
    i just want to confirm if my answer is right....
    this is the problem:

    solve the expression for n Є N

    P(2n + 4, 3) = 2/3P(n+4, 4)

    this is my work:

    (2n +4)!/((2n +4)-3)!) = (2/3(n+4)!)/((n+4)!)-4)!)
    (2n +4)!/((2n +1)! )= (2/3(n+4)!)/(n!)

    and i dont know wat to do next... should i just cross multiply it?
     
    Last edited: Nov 5, 2005
  2. jcsd
  3. Nov 5, 2005 #2
    permutations

    help me with this if my answer is correct....
    help is appreciated...
     
    Last edited: Nov 5, 2005
  4. Nov 6, 2005 #3
    help from anyone with this problem...
     
    Last edited: Nov 6, 2005
  5. Nov 6, 2005 #4
    start canceling terms. if you have [tex]\frac{(2n+4)!}{(2n+1)!}[/tex] and you understand what factorial means, you can cancel all the factors in the denominator with some in the numerator. do this as well to the right side of your equation. then try to solve.
     
  6. Nov 6, 2005 #5
    so you mean [tex]\frac{(2n+4)!}{(2n+1)!}[/tex] is same thing as writing as (2n+4)(2n+3)(2n+2)(2n+1)/(2n+1)?
     
  7. Nov 6, 2005 #6
    uh, well the (2n+1)'s cancel out. but ya. thats how a factorial works. so do the same to other side and go.
     
  8. Nov 6, 2005 #7
    correct me if im wrong ok? i've try the other side... should you make the numerator 2/3(n+4)! to -2/3(n-4)! in order to cancel it from n(n-1)(n-2)(n-3)(n-4)?
     
  9. Nov 6, 2005 #8
    no... i think you're missing something...

    [tex]\frac{2/3(n+4)!}{n!}= \frac{2/3(n+4)(n+3)(n+2)(n+1)(n)(n-1)(n-2)(n-3).....}{n(n-1)(n-2)(n-3)......}[/tex]
    so you're left with only [tex] 2/3(n+4)(n+3)(n+2)(n+1)[/tex] cause you can cancel the rest out. do you see this?? does it make sense? this is what we did to the other side as well.
     
  10. Nov 6, 2005 #9
    now i get gale... thanks so m uch, and it is clearer for me now...
     
  11. Nov 6, 2005 #10
    what if you get like this.... -24(2n+3)(n+4)(n+3) = 0...
    what should i do to the -24? or it is just 0, then the value for n is just n=-2/3, n=-4 and n=-3?
     
  12. Nov 6, 2005 #11
    well, i'm not really sure how you got that.... its not what i get. what's your equation after you've simplified? i get something simple enough to multiply out and then combine like terms. then solve.
     
  13. Nov 6, 2005 #12
    i got like (2n+4)(2n+3)(2n+2) = 2/3(n+4)(n+3)(n+2)(n+1)
    then... 2(n+2)(2n+3)2(n+1) = 2/3(n+4)(n+3)(n+2)(n+1)
    which is 4(2n+3) = 2/3(n+4)(n+3)
    and i dont know wat to do next
     
  14. Nov 6, 2005 #13
    multiply it out and simplify, its just a quadractic.
     
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