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Math Help - Permutations and Combinations

  1. #1
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    Permutations and Combinations

    Q1: The number of ways 8 people can be arranged in a straight line if two people insist on being separated is:
    A) 15,120
    B) 3,600
    C) 30,240
    D) 7,200
    E) 5,400


    No matter what I try I can't obtain any of these possible answers. I'm thinking it's something to do with the wording of the question, but I'm not sure what exactly.


    Q2: A cricket team takes 11 players to a game and there are 14 players from which to choose; 1 wicket keeper, 7 batsmen and 6 bowlers.
    a) How many different teams are possible if there are no restrictions?

    I did: 14C11 = 364, but I'm unsure.

    b) How many different teams are possible if the wicket keeper must be included?
    I did 13C11 = 78, but I'm unsure.

    c) How many different teams are possible if the team contains the wicket keeper, 6 batsmen and 4 bowlers?


    Thanks in advance.
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  2. #2
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    Re: Permutations and Combinations

    Hello, Fratricide!

    Q1: the number of ways 8 people can be arranged in a straight line
    . . . if two people insist on being separated is:

    . . (A) 15,120 . . (B) 3,600 . . (C) 30,240 . . (D) 7,200 . . (E) 5,400

    \text{The eight people are: }\:\{A,B,C,D,E,F,G,H\}

    \text{Let }A\text{ and }B\text{ be the two unfriendly people.}

    \text{With no restrictions, there are }\,8! = 40,\!320\text{ arrangements.}


    \text{Now consider the arrangements in which }A\text{ and }B\:are\text{ adjacent.}

    \text{Duct-tape }A\text{ and }B\text{ together.}
    \text{Then we have 7 "people" to arrange: }\,\left\{\boxed{AB}, C, D, E, F, G, H\right\}
    \text{There are: }\,7!\text{ arrangements.}

    \text{But }A\text{ and }B\text{ could be taped like this: }\,\boxed{BA}.

    \text{Hence, there are: }\,2\cdot7! \,=\,10,\!080\text{ ways in which }A\text{ and }B\text{ are adjacent.}


    \text{Therefore, the answer is: }\:40,\!320 - 10,\!080 \:=\:30,\!240\:\text{ . . . answer (C)}

    Thanks from Fratricide
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  3. #3
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    Re: Permutations and Combinations

    Hello
    I think you are right in Q2(a)

    (b) 13C10( because the wicket keeper must be include)
    (c) 7C6x6C4
    Thanks from Fratricide
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    Re: Permutations and Combinations

    Quote Originally Posted by Trefoil2727 View Post
    Hello
    I think you are right in Q2(a)

    (b) 13C10( because the wicket keeper must be include)
    (c) 7C6x6C4
    Could you please elaborate on (c) to clarify my understanding?
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  5. #5
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    Re: Permutations and Combinations

    Quote Originally Posted by Fratricide View Post
    Could you please elaborate on (c) to clarify my understanding?
    well I really not good on explanation, because my English is sucked, but I'll try..
    you have to choose 11 players, and there is only 1 wicket keeper, which you have to count in. So there left only 13 people to choose, as the question ask to have 6 batsmen( from 7) and 4 bowlers( from 6). Therefore, 7C6x6C4
    Thanks from Fratricide
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  6. #6
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    Re: Permutations and Combinations

    Quote Originally Posted by Trefoil2727 View Post
    well I really not good on explanation, because my English is sucked, but I'll try..
    you have to choose 11 players, and there is only 1 wicket keeper, which you have to count in. So there left only 13 people to choose, as the question ask to have 6 batsmen( from 7) and 4 bowlers( from 6). Therefore, 7C6x6C4
    Perfect, thanks.
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