Results 1 to 7 of 7

Thread: combinations equation.

  1. #1
    Member
    Joined
    Aug 2017
    From
    England
    Posts
    91
    Thanks
    1

    combinations equation.

    x nCr 5 = 4[(x-1) nCr 4]


    does anyone know how to solve for x here?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Jun 2013
    From
    Lebanon
    Posts
    855
    Thanks
    402

    Re: combinations equation.

    use

    $$\binom {x} {5} = \binom {x - 1} {5} + \binom {x - 1} {4}$$

    and simplify
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Nov 2010
    Posts
    3,634
    Thanks
    1458

    Re: combinations equation.

    Quote Originally Posted by edwardkiely View Post
    x nCr 5 = 4[(x-1) nCr 4]


    does anyone know how to solve for x here?
    Use the definition for ${_n C_r} = \dfrac{n(n-1)(n-2)\cdots (n-r+1)}{r!}$

    $\dfrac{x(x-1)(x-2)(x-3)(x-4)}{5!} = 4\dfrac{(x-1)(x-2)(x-3)(x-4)}{4!}$

    Cancel out $(x-1)(x-2)(x-3)(x-4)$ on both sides, multiply both sides by $5!$, you wind up with $x=4\cdot 5 = 20$.
    Last edited by SlipEternal; Mar 8th 2018 at 04:29 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Nov 2010
    Posts
    3,634
    Thanks
    1458

    Re: combinations equation.

    Or, obviously, x=1,2,3, or 4, so long as you are using the definition that ${_nC_r} = 0$ if $0\le n<r$
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Aug 2017
    From
    England
    Posts
    91
    Thanks
    1

    Re: combinations equation.

    we are supposed to be able to solve it just using the the formula n!/r!*(n-r)!

    do you know how i can solve just using this
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Nov 2010
    Posts
    3,634
    Thanks
    1458

    Re: combinations equation.

    Quote Originally Posted by edwardkiely View Post
    we are supposed to be able to solve it just using the the formula n!/r!*(n-r)!

    do you know how i can solve just using this
    Keep in mind the following:
    $n! = n\cdot (n-1)! = n(n-1)\cdot (n-2)!$

    So, $n! = n(n-1)(n-2)\cdots (n-r+1)(n-r)!$

    Then your equation becomes:

    ${_nC_r} = \dfrac{n!}{r!(n-r)!} = \dfrac{n(n-1)(n-2)\cdots (n-r+1) \cancel{(n-r)!}}{r!\cancel{(n-r)!}} = \dfrac{n(n-1)(n-2)\cdots (n-r+1)}{r!}$

    Then see post #3.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Aug 2017
    From
    England
    Posts
    91
    Thanks
    1

    Re: combinations equation.

    thanks

    i get it know
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. How many combinations?
    Posted in the Statistics Forum
    Replies: 1
    Last Post: Aug 2nd 2011, 02:33 PM
  2. Combinations of a Set
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: Jan 17th 2011, 06:08 AM
  3. combinations
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: Jan 15th 2011, 04:38 AM
  4. Combinations
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: Jan 25th 2010, 09:00 PM
  5. Combinations
    Posted in the Advanced Statistics Forum
    Replies: 6
    Last Post: May 6th 2006, 05:10 PM

/mathhelpforum @mathhelpforum