Results 1 to 4 of 4

Math Help - Inclusion Exclusion

  1. #1
    Junior Member
    Joined
    Nov 2009
    Posts
    51

    Inclusion Exclusion

    How many ways to arrange the letters of SCIENCE BRAIN
    such that there are exactly 3 sets of the same letter conseculatively.

    the total number of ways when doing regularily is 14!/2^4

    since the question says that it wants 3 sets of the same letter conseculatively im thinking i would find that there are 13!/2^3 ways to have 1 set of the same letter consecutatively...except that just blocks one set of letters and doesnt garuntee for the other letters..and normally when doing PIE the cases are the opposite of the question so should i be finding the ways such that they are not consecutive?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,677
    Thanks
    1618
    Awards
    1
    Quote Originally Posted by stumped765 View Post
    How many ways to arrange the letters of SCIENCE BRAIN
    such that there are exactly 3 sets of the same letter conseculatively.
    the total number of ways when doing regularily is 14!/2^4
    There are four ways to select three pairs to be together.
    Then there are \dfrac{11!}{2} ways to have those three pairs together.
    But that also counts the case in which all four pairs are together. Thus subtract that case.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Nov 2009
    Posts
    51
    Quote Originally Posted by Plato View Post
    There are four ways to select three pairs to be together.
    Then there are \dfrac{11!}{2} ways to have those three pairs together.
    But that also counts the case in which all four pairs are together. Thus subtract that case.
    if we had sCCIINNeebr thats 3 pairs, so the number of ways to have that.. wouldnt that be 8!/2? since the CC II NN act as blocks?

    This is what i have
    4C3*8!/2 - 4C4*7!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,677
    Thanks
    1618
    Awards
    1
    Quote Originally Posted by stumped765 View Post
    if we had sCCIINNeebr thats 3 pairs, so the number of ways to have that.. wouldnt that be 8!/2? since the CC II NN act as blocks? This is what i have 4C3*8!/2 - 4C4*7!
    I took your own count. But there are 12 letters not 14.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Inclusion/Exclusion
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: January 19th 2011, 03:49 AM
  2. inclusion & exclusion help
    Posted in the Discrete Math Forum
    Replies: 11
    Last Post: May 14th 2010, 09:57 AM
  3. Inclusion/Exclusion
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: November 3rd 2009, 08:46 AM
  4. Inclusion-exclusion
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: July 28th 2009, 02:45 AM
  5. inclusion/exclusion
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: November 18th 2008, 10:31 PM

Search Tags


/mathhelpforum @mathhelpforum