Results 1 to 5 of 5

Math Help - Combinations

  1. #1
    Senior Member
    Joined
    Oct 2009
    Posts
    459

    Combinations

    My answer to this question is 2 X 5! = 240. The book gives 1440, presumably 2 X 6!.

    A concert pianist has prepared seven different pieces of music for recital, three of which are modern and four are classical. Calculate the number of different orders in which she can play the seven pieces when the recital must start and end with classical pieces.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor undefined's Avatar
    Joined
    Mar 2010
    From
    Chicago
    Posts
    2,340
    Awards
    1
    Quote Originally Posted by Stuck Man View Post
    My answer to this question is 2 X 5! = 240. The book gives 1440, presumably 2 X 6!.

    A concert pianist has prepared seven different pieces of music for recital, three of which are modern and four are classical. Calculate the number of different orders in which she can play the seven pieces when the recital must start and end with classical pieces.
    I get there via 4*(6!-3*5!), alternatively, 4*3*5!, alternatively, P(4,2)*5!
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Oct 2009
    Posts
    459
    P(4,2)*5! is a valid method. The end pieces are the permutations of any 2 from 4 classical pieces. The centre pieces are the permutations of the 5 remaining pieces.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor undefined's Avatar
    Joined
    Mar 2010
    From
    Chicago
    Posts
    2,340
    Awards
    1
    Quote Originally Posted by Stuck Man View Post
    P(4,2)*5! is a valid method. The end pieces are the permutations of any 2 from 4 classical pieces. The centre pieces are the permutations of the 5 remaining pieces.
    Yes and the other two methods are also valid.

    4*(6!-3*5!) -- 4 ways to choose the first piece, then 6! permutations of the rest, but 3*5! of those don't end with classical.

    4*3*5! -- 4 ways to choose first piece, 3 ways to choose last piece, 5! ways to permute the rest.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Alternatively, the central 5 contain \binom{4}{2} pairs of classical pieces, along with the 3 modern pieces.

    There are 5! arrangements of the central 5 and for each of these, there are 2! arrangements
    of the first and last classical pieces.

    That's \binom{4}{2}5!2!

    Or...

    There are \binom{4}{2} ways to select the 1st and last classical pieces.

    There are 5! arrangements of the central 5 pieces and for each of these there are 2! arrangements
    of the outer 2 pieces....again \binom{4}{2}5!2!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Combinations in a set
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: September 9th 2010, 06:19 AM
  2. How many combinations are possible?
    Posted in the Discrete Math Forum
    Replies: 0
    Last Post: July 23rd 2009, 07:53 PM
  3. Combinations
    Posted in the Statistics Forum
    Replies: 2
    Last Post: May 5th 2008, 08:28 AM
  4. How many combinations..?
    Posted in the Algebra Forum
    Replies: 9
    Last Post: May 2nd 2008, 10:34 AM
  5. combinations
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: April 27th 2008, 09:00 AM

Search Tags


/mathhelpforum @mathhelpforum