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Math Help - Arrangements

  1. #1
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    Exclamation Arrangements

    1. A commitee of seven is to be selected from 10 teachers and 12 students. How many different committees are possible if there is to be a majority of students with at least one teacher on the committee?

    2. Sarah, Vanessa, jessica, David, Paul and Mark are seated at a round table. How many different arrangements are possible if the boys and girls must alternate and Jessica can not sit beside Paul.

    Ah, permutations and combinations - my worst weakness. Could someone please explain to me how to do these sort of questions?
    Last edited by mr fantastic; September 5th 2009 at 09:53 PM.
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  2. #2
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    Solution using Multiplication Rule and Combination Rule

    Quote Originally Posted by xwrathbringerx View Post
    1. A commitee of seven is to be selected from 10 teachers and 12 students. How many different committees are possible if there is to be a majority of students with at least one teacher on the committee?
    I believe this is the solution.
    You have to the sum of all combinations (order doesn't matter) of the following combinations (excuse the redundancy).
    A = 1 teacher & 6 students
    B = 2 teachers & 5 students
    C = 3 teachers & 4 students

    The combination rule: (n!) [(n-k)!n!]. I am assuming further explanation regarding the factorial "!" is in your textbook.

    Let C(A) represent the number of combinations of A, and so on.
    C(A) = (7 choose 1) * (12 choose 6) = 7x924.
    Do this for the remaining two possible general combinations, B and C. Then sum the total combinations to get 40425.

    I hope this helps. Let me know if you have any questions.

    Best of luck!
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  3. #3
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    Quote Originally Posted by xwrathbringerx View Post
    2. Sarah, Vanessa, jessica, David, Paul and Mark are seated at a round table. How many different arrangements are possible if the boys and girls must alternate and Jessica can not sit beside Paul.

    Ah, probability - my worst weakness. Could someone please explain to me how to do these sort of questions?
    Try this solution:

    They are asking all the possible arrangements (order matters) minus those arrangements where Jessica is next to Paul at a round table.

    All the possible arrangements for boy-girl-boy-girl-boy-girl seating:
    In the first seat you have 3 boys to choose from, the next seat 3 girls to choose from, the next seat 2 boys to choose from because you have already pick one for the first seat, the next seat 2 girls to choose from, then you have 1 boy and 1 girl left for the remaining seats. **Use the Factorial to put n different objects in order. It's slightly altered in this case because of the alternating condition.***

    3x3x2x2x1x1 = 36 total possible arrangements

    Now you calculate those arrangements where Jessica is next to Paul. A picture would be best for this explanation but I will try my best to explain in words anyway.

    Let Jessica sit first then Paul sit next. So we have 1 x 1. We multiply this with the remaining combinations for the other 4 seats. Using the same logic as above the remaining four seats will have 2x2x1x1 = 4 arrangements.

    36 total arrangements - 4 arrangements not allowed = 32 options.
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  4. #4
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    Quote Originally Posted by xwrathbringerx View Post
    2. Sarah, Vanessa, Jessica, David, Paul and Mark are seated at a round table. How many different arrangements are possible if the boys and girls must alternate and Jessica can not sit beside Paul.
    Surprisingly the answer to this question is just 4.
    Remember these are circular arrangements. With no restrictions there are 5!=120 ways to seat six people at a circular table.
    If we insist that the boys and girls must alternate, then there are 2!\cdot 3!=12 ways to seat this group.
    But if we add the condition ‘Jessica can not sit beside Paul, then there are only 4 ways to do it.
    To see why, seat Paul at the table. There are two choices of a girl to seat at Paul’s right. Now there are two choices of a boy to seat at that girl’s right. At this point there is only one way to seat the remaining three people.
    Last edited by Plato; September 6th 2009 at 07:22 AM.
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  5. #5
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    Quote Originally Posted by xwrathbringerx View Post
    1. A commitee of seven is to be selected from 10 teachers and 12 students. How many different committees are possible if there is to be a majority of students with at least one teacher on the committee?

    2. Sarah, Vanessa, jessica, David, Paul and Mark are seated at a round table. How many different arrangements are possible if the boys and girls must alternate and Jessica can not sit beside Paul.

    Ah, permutations and combinations - my worst weakness. Could someone please explain to me how to do these sort of questions?

    First Problem

    The condition in the first problem can be restated as follows:

    A committee of seven comprises at least 4 students and at least 1 teacher. Given the constraints, a valid committee would comprise of 4 students and 3 teachers, or 5 students and 2 teachers, or 6 students and 1 teacher. So, the total number of ways in which a committee of seven can be formed is the sum total of number of ways in which in which a committee of seven is formed with x number of students, and y number of teachers, where x ranges from 4 to 6 while y ranges from 3 to 1.

    The total number of ways in which a committee of seven can be formed given there are x students and y teachers equals the number of ways we can select x students from 12 students and y teachers from 10. (Basic multiplication principle)

    Needless to say you'd use the combination formula for evaluating the number of ways you select a certain number of elements from a certain number of distinguishable elements.

    Second Problem

    You can try solving the problem for a simpler case first, and then proceed to a more difficult case.

    Here's my attempt: Using the analogy of filling balls in boxes and first solving for the case of linear arrangement (not a circular one), we have 6 boxes (positions) to fill with 6 balls (persons) such that three distinguishable red balls (males) and three distinguishable white balls (girls) alternate, and a certain red ball, say R1 (Paul) and a certain white ball W1 (Jessica) are never placed adjacent to each other.

    Let us assume that first box is to be filled with a red ball. Then following the condition we will have the following arrangement: RWRWRW. By doing so, we have, thus, fixed the positions for red and white balls. The first position for red balls can be filled by any of the 3 red balls, while the second can be filled by any of the two, and for the third we have only one choice. Thus, we have, all in all, 3*2*1 different ways in which we can place of our red balls. Now we'll find the possible number of arrangements of white balls for any given arrangement of red balls. For that sake, we have in all 4 different arrangements of white balls for each arrangement of red balls. We can't have six different arrangements of white balls because we are restricted not to place Jessica and Paul side by side. And we can impose that restriction by allowing Paul to choose his place freely while restricting Jessica not to sit with him. There are two ways for Jessica to sit besides Paul, either on his left or his right, for any given place where Paul is allowed to sit. Thus, we have four different valid arrangements for each arrangement of red balls. Our total number of arrangements, thus, become 24. (repeated use of the Basic Multiplication Principle)

    For our purposes, we don't need to consider the scheme of seating in which a white ball is placed in the first box as we are interested in finding a solution for the case of circular arrangement. In a circular arrangement, we don't distinguish between the following linear arrangements: ABC, CAB, BCA. The reason being these arrangements look the same when placed along a circle's boundary. Thus in our case for 6 such linear arrangements of white and red balls we have one circular arrangement.

    So we are going to divide total number of linear arrangements (24) by 6 to yield four.

    I hope my reasoning is correct.
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