Results 1 to 5 of 5

Math Help - combination letters with restrictions

  1. #1
    Newbie
    Joined
    Jan 2009
    Posts
    5

    combination letters with restrictions

    How many arrangements are there for PROBABILITIES if there are

    - no 2 vowels together

    - there is no E next to an I
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,959
    Thanks
    1783
    Awards
    1
    b) These are the consonants are: P,R,B,B,L,T,S.
    They can be arranged in \frac{7!}{(2!)} ways;
    Each of the ways creates 8 places _P_R_B_B_L_T_S_ to put the vowels.
    The number of places is then 8 \choose 6. The number of vowel arrangements is \frac{6!}{(3!)}
    Put it together: {\frac{7!}{(2!)}} {8 \choose 6} \frac{6!}{(3!)}.

    a) “PROBABILITIES” the word has 13 letters but some repeat.
    There are \frac{13!}{(3!)(2!)} ways to arrange this word: 3 I’s & 2 B’s.
    You now must find the number of ways we can have “EI”, “IE”, or “IEI”
    And subtract from total.
    Last edited by mr fantastic; January 26th 2009 at 11:25 AM. Reason: No edit - just flagging this reply as having been moved from a duplicate thread.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Thanks
    1

    Arrangements with repeated letters

    Hello PatCal
    Quote Originally Posted by PatCal View Post
    How many arrangements are there for PROBABILITIES if there are

    - no 2 vowels together

    - there is no E next to an I
    The standard method for solving letter re-arrangements is to assume first that any repeated letters are distinguishable - so here we would call the two B's B_1 and B_2, and the three I's I_1, I_2 and I_3. Then work out the number of arrangements with all the letters different. Here that would be 13!.

    Finally note that because of repeated letters, there will be duplications. So we divide this total by the number of arrangements of each group of repeated letters within themselves. In the case of the 2 B's, that's 2!, and for the 3 I's, it's 3!. So the answer would be

    \frac{13!}{2!3!}

    But we have extra restrictions, so we shall need to apply additional techniques.

    So, for question 1, where no two vowels are together. There are six vowels: O, A, E and three I's; and seven consonants. If no two vowels can come together, and we number the positions 1 to 13, then the vowels must occupy positions 2, 4, 6, 8, 10 and 12; and the consonants the odd-numbered positions.

    So, the six vowels could be arranged in their positions in 6! ways if they were all different. But, because of the duplication among the three I's we need to divide this number by 3!, the number of ways of arranging the I's among themselves. Thus the number of ways in which the vowels can be placed is

    \frac{6!}{3!}

    Now do the consonants in a similar way, remembering to divide to take into account the duplication of the two B's.

    Finally, multiply the number of arrangements of the vowels by the number of arrangements of the consonants to get the total.

    For question 2, where there is no E next to an I. This is most easily solved by finding the number of arrangements in which an E and an I will come together and subtracting this from the overall total. So, how many ways are there of combining an I and an E into a single letter-group? Clearly, there are just two: IE or EI.

    Now treat this letter group as a single character, and work out how many ways there now are of arranging this in a line together with the other eleven letters, remembering to take into account the duplication that will arise because of the 2 B's and the remaining 2 I's. *See PS

    Subtract this from \frac{13!}{2!3!} and you're done.

    Can you do it from here?

    Grandad

    *PS. Sorry, there's an additional factor to take into account. That is, the number of times the arrangement IEI occurs, because each of these will be counted twice in the above method: once where the IE group is placed in the line, and then has an I placed on its right, and once with the EI group being placed and then an I on its left.

    So, you need to work out how many times you'll get the combination IEI occurring, and add this back in - because we shall have taken it away twice, when once would have been correct. This number, then, is the number of arrangements of the letter-group IEI and the remaining ten letters - which this time only have B as a duplicated letter. Number of arrangements containing IEI, then, is

    \frac{11!}{2!}

    and I make the final answer to question 2: 299,376,000

    Tricky!
    Last edited by Grandad; January 26th 2009 at 09:39 AM. Reason: Correction to final answer
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,907
    Thanks
    766
    Hello, PatCal!

    I think I have the first part.
    (I'm still working on the second part.)


    How many arrangements are there for PROBABILITIES if there are

    (a) no two vowels together

    \text{We have: }\:\underbrace{\{A,E,I,I,I,O\}}_{\text{6 vowels}} \cup \underbrace{\{B,B,L,P,R,S,T\}}_{\text{7 consonants}}


    Place the 7 consonants in a row, leaving a space before, after, and between them.

    . . \begin{array}{ccccccccccccccc}\_ & C & \_ & C & \_ & C & \_ & C & \_ & C & \_ & C & \_ & C & \_ \end{array}

    There are: . \frac{7!}{2!} orderings for the consonants.


    Place the 6 vowels in the spaces.
    . . There are: . {8\choose6} choices for the six spaces.
    . . and there are: . \frac{6!}{3!} orderings for the vowels.


    Therefore: . \frac{7!}{2!}\cdot{8\choose6}\cdot \frac{6!}{3!} \:=\:2520 \cdot 28 \cdot 120 \:=\:8,\!467,\!200

    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Thanks
    1

    Arrangements with repeated letters

    Hello PatCal and Soroban -

    My first posting was incorrect. My apologies. In question 1, the vowels don't have to be placed in positions 2, 4, 6, 8, 10 and 12, of course. They could go in any six of the eight spaces, as Soroban has said!

    Grandad
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Combinations with restrictions.
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: July 20th 2011, 09:49 PM
  2. Restrictions for logarithms
    Posted in the Algebra Forum
    Replies: 7
    Last Post: January 5th 2011, 03:23 PM
  3. combination's of letters
    Posted in the Statistics Forum
    Replies: 4
    Last Post: May 24th 2009, 01:17 AM
  4. restrictions
    Posted in the Algebra Forum
    Replies: 2
    Last Post: December 4th 2008, 11:11 AM
  5. restrictions for x
    Posted in the Algebra Forum
    Replies: 2
    Last Post: July 27th 2008, 01:33 PM

Search Tags


/mathhelpforum @mathhelpforum