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Thread: committee

  1. #1
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    committee

    a committee contains 4 english and 7 french and 5 german representatives . three are chosen at random from a sub committee . find the probability that all three are different nationalities.

    i tried this by finding the probability of an english,french and german person picked mulitplied by three are there is 3 ways to arrange this but it did not give me the correct answer
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  2. #2
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    Re: committee

    $\dfrac{4\cdot 7\cdot 5}{\dbinom{16}{3}} = \dfrac{1}{4}$
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  3. #3
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    Re: committee

    A deck of 16 cards:
    4 cards labelled 1, 5 cards labelled 2, 7 cards labelled 3.

    Deck is shuffled, then 3 cards picked at random without replacement.

    What is probability that the 3 cards have different numbers?

    Marko, agree that above is same as your problem?
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  4. #4
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    Re: committee

    thanks that is the right answer . could you explain how you got this. ? have you written down the short version of the answer.?

    i understand 16nCr3 is the total number of ways 3 people can be picked. what does the 4*7*5 stand for
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  5. #5
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    Re: committee

    Quote Originally Posted by markosheehan View Post
    thanks that is the right answer . could you explain how you got this. ? have you written down the short version of the answer.?

    i understand 16nCr3 is the total number of ways 3 people can be picked. what does the 4*7*5 stand for
    Choose an English representative: $\dbinom{4}{1} = 4$
    Choose a French representative: $\dbinom{7}{1} = 7$
    Choose a German representative: $\dbinom{5}{1} = 5$

    By the product principle, the number of ways to choose one of each is $4\cdot 7\cdot 5$.
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  6. #6
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    Re: committee

    the order does not matter though so saying 5*7*4 and saying this is the total number of ways you can pick one person from each nationality would mean the order does matter.
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  7. #7
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    Re: committee

    Quote Originally Posted by markosheehan View Post
    the order does not matter though so saying 5*7*4 and saying this is the total number of ways you can pick one person from each nationality would mean the order does matter.
    Does order matter when one does multiplication?

    Is it true that $5\cdot 7\cdot 4=4\cdot 5\cdot 7=7\cdot 5\cdot 4~?$
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  8. #8
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    Re: committee

    Quote Originally Posted by markosheehan View Post
    the order does not matter though so saying 5*7*4 and saying this is the total number of ways you can pick one person from each nationality would mean the order does matter.
    Saying that order does matter would mean that you need to multiply by 3! (the number of ways to order the three candidate picks).

    Another way to think of it. The whole set looks like this:

    $\{E_1,E_2,E_3,E_4,F_1,F_2,F_3,F_4,F_5,F_6,F_7,G_1 ,G_2,G_3,G_4,G_5\}$

    We want to find the number of three-element subsets of the form:

    $\{E_i,F_j,G_k\}$

    Well, $i=1,2,3,4$, $j=1,2,3,4,5,6,7$, and $k=1,2,3,4,5$. When I write a subset, I do not consider the order that I write the elements. I just make sure the elements are there. Still, I have 4 choices for the E, 7 choices for the F, and 5 choices for the G. That is $4\cdot 7\cdot 5$ possible subsets.
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  9. #9
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    Re: committee

    thanks for your replies i get it now
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