Results 1 to 4 of 4
Like Tree2Thanks
  • 1 Post By emakarov
  • 1 Post By Plato

Math Help - Combinations

  1. #1
    Junior Member Greymalkin's Avatar
    Joined
    Jun 2012
    From
    Montreal
    Posts
    74
    Thanks
    1

    Combinations

    2 more problems:
    Combinations-trig2.png
    I insert the above in brackets into the combination formula nCk
    which becomes: \frac {n!}{(n-2)!(n-(n-2))!}=6

    solving for n I get 1.12 or 3.33, book gives the answer of 4.

    I am incorrect for the second question(45) as well using the same method, my answer is 0/1/-7/-8/-9(approximated), book answer is 7.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Oct 2009
    Posts
    5,536
    Thanks
    778

    Re: Combinations

    Can you show how you get your answers? The first problem reduces to a quadratic equation. The second problem reduces to an equation of fourth degree, but after cancellation it also becomes a quadratic equation.
    Thanks from Greymalkin
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member Greymalkin's Avatar
    Joined
    Jun 2012
    From
    Montreal
    Posts
    74
    Thanks
    1

    Re: Combinations

    Quote Originally Posted by emakarov View Post
    Can you show how you get your answers? The first problem reduces to a quadratic equation. The second problem reduces to an equation of fourth degree, but after cancellation it also becomes a quadratic equation.
    I put both sides of the equations into a graphing calculator and use the intersections as "n".

    (n!)/((n-2)!(n-(n-2))!)=6 - Wolfram|Alpha

    Looking back on my cellphones wolfram application history it was a transcription error after all.(for 43)

    45:
    ((n+2)!/(4!(&# 40;n-2)!))=6((n!)&# 41;/(2!(n-2)!) - Wolfram|Alpha
    apologies
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,649
    Thanks
    1597
    Awards
    1

    Re: Combinations

    Quote Originally Posted by Greymalkin View Post
    2 more problems:
    Click image for larger version. 

Name:	trig2.png 
Views:	2 
Size:	8.0 KB 
ID:	24924
    I insert the above in brackets into the combination formula nCk
    which becomes: \frac {n!}{(n-2)!(n-(n-2))!}=6
    \binom{n+2}{4}=\frac{(n+2)(n+1)(n)(n-1)}{4!} and 6\binom{n}{2}=6\frac{n(n-1)}{2!}~.

    Put those equal we get n^2+3n-70=0~.
    Last edited by Plato; September 25th 2012 at 11:45 AM.
    Thanks from Greymalkin
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. combinations
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: January 15th 2011, 04:38 AM
  2. How many combinations?
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: April 5th 2010, 08:26 AM
  3. Combinations
    Posted in the Algebra Forum
    Replies: 3
    Last Post: May 20th 2009, 04:03 PM
  4. combinations
    Posted in the Statistics Forum
    Replies: 1
    Last Post: September 9th 2008, 03:44 PM
  5. combinations
    Posted in the Advanced Statistics Forum
    Replies: 11
    Last Post: September 4th 2008, 10:26 PM

Search Tags


/mathhelpforum @mathhelpforum