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Math Help - Permutations and combinations! Help needed!

  1. #1
    Newbie IceDancer91's Avatar
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    Permutations and combinations! Help needed!

    Can sumbuddy plz help me with Q6 part ii and iii of the attached paper?? I have attached the marking scheme as well- even though i have the solution, i am still not able to understand how they are doing it... Can somebody plz guide me on that? thanks in advance! =)
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    Member kalyanram's Avatar
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    Re: Permutations and combinations! Help needed!

    Q: A small aeroplane has 14 seats for passengers. The seats are arranged in 4 rows of 3 seats and a back
    row of 2 seats (see diagram). 12 passengers board the aeroplane.
    (i) How many possible seating arrangements are there for the 12 passengers? Give your answer
    correct to 3 significant figures. [2]

    These 12 passengers consist of 2 married couples (Mr and Mrs Lin and Mr and Mrs Brown), 5 students
    and 3 business people.
    (ii) The 3 business people sit in the front row. The 5 students each sit at a window seat. Mr and Mrs
    Lin sit in the same row on the same side of the aisle. Mr and Mrs Brown sit in another row on
    the same side of the aisle. How many possible seating arrangements are there? [4]
    (iii) If, instead, the 12 passengers are seated randomly, find the probability that Mrs Lin sits directly
    behind a student and Mrs Brown sits in the front row. [4]

    Sol:
    (ii)
    With respect to the (Figure-1) attached.
    The three businessmen can choose from the first row (indicated in blue) ways they can be arranged = 3!
    The married couple can choose from the paired seat (indicated in green) the first pair getting to choose has 3 choice the second pair has 2 choices and the pair can be considered as Mr, Mrs or Mrs, Mr so total arrangements = ( 3x 2)x( 2x 2)

    With respect to the (Figure-2)
    Now the students get to choose among the remaining 5 window seats (one from the dotted red box, 4 from the solid red box) now the number of ways they can be arranged = 5!

    \therefore Total no.of arrangements = 3!x( 3x 2)x( 2x 2)x 5!

    (iii) Can be solved on similar line of grouping the cases.

    ~Kalyan.
    Attached Thumbnails Attached Thumbnails Permutations and combinations! Help needed!-ap.png  
    Last edited by kalyanram; September 5th 2012 at 06:24 AM.
    Thanks from IceDancer91
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    Newbie IceDancer91's Avatar
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    Re: Permutations and combinations! Help needed!

    Can u plz solve the last part too? :/
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  4. #4
    Member kalyanram's Avatar
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    Re: Permutations and combinations! Help needed!

    Show me your attempt and the difficulty your are facing. I will surely help you.

    Kalyan.
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    Newbie IceDancer91's Avatar
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    Re: Permutations and combinations! Help needed!

    Umm ohk. I somehow sorted that out my self.

    Mrs Brown's probability of sitting in the front seat is 3/10. Out of the remaining 13 seats (1 front seat is occupied by Mrs Brown)- there are 10 such seats in which Mrs Lin can place herself so that she will be directly behind a student, so her probability of sitting behind a student is 10/13 and there are 5 students so their probability is 5/12. (Sorry there's a stupid question i have in mind- is this 12 in the denominator is the total number of passengers i.e. 12 or the remaining seats after 2 are occupied?)

    So P= 3/10 * 10/13* 5/12= .0687

    Using another method, Mrs Brown can sit in 3 front seats, Mrs Lin in 10, and 5 students will obviously occupy 5 seats, the rest of 9 passengers can take 11 seats.
    So P= (3*10*5*11P9)/ 14P12

    Is what i am understanding out of the solution correct? or there is still a problem? I really hate permutations and combinations- seriously.

    P.S. Sorry, i attached the wrong marking scheme before. The correct one is attached with this post.
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  6. #6
    Member kalyanram's Avatar
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    Re: Permutations and combinations! Help needed!

    Quote Originally Posted by IceDancer91 View Post
    Mrs Brown's probability of sitting in the front seat is 3/10. ..... So P= 3/10 * 10/13* 5/12= .0687.
    This is incorrect the probability of Mrs. Brown is \frac{3}{14} hence over all probability = \frac{3}{14}.\frac{10}{13}.\frac{5}{12}

    Quote Originally Posted by IceDancer91 View Post
    Sorry there's a stupid question i have in mind- is this 12 in the denominator is the total number of passengers i.e. 12 or the remaining seats after 2 are occupied?
    The probability is with respect to the seat selection after two are occupied.

    Quote Originally Posted by IceDancer91 View Post
    5 students will obviously occupy 5 seats
    This argument is incorrect. The correct argument goes like this you have picked 3 suitable places for Mrs. Brown and 10 suitable places for Mrs.Lin. Now your job is to place one student in front of Mrs.Lin; and the choices you have are \binom{5}{1}=5 and the remaining people can pick the seats in {^{11}P_9} ways. And as you have mentioned total possible arrangements are {^{14}P_{12}}. Probability of the event is \frac{3.10.5.{^{11}P_9}}{{^{14}P_{12}}}

    Kalyan.
    Last edited by kalyanram; September 6th 2012 at 12:31 AM.
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    Newbie IceDancer91's Avatar
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    Re: Permutations and combinations! Help needed!

    OH OHK. Thanks a lot for your help. =)
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