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Math Help - Combinations in combinations?

  1. #1
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    Combinations in combinations?

    Given are all the possible 10 number combinations (NOT permutations) of the numbers 11 - 26 (8008 combinations), and 4 random numbers from the 11 - 26 sequence.
    How do I find out how many of the 8008 combinations contain
    a. all 4 random numbers?
    b. 3 of the 4, 2 of the 4, 1 of the 4 and 0 of the 4 random numbers respectively??
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    Quote Originally Posted by aimpro2000 View Post
    Given are all the possible 10 number combinations (NOT permutations) of the numbers 11 - 26 (8008 combinations), and 4 random numbers from the 11 - 26 sequence.
    How do I find out how many of the 8008 combinations contain
    a. all 4 random numbers?
    b. 3 of the 4, 2 of the 4, 1 of the 4 and 0 of the 4 random numbers respectively??
    What do you mean by random number?
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  3. #3
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    any 4 numbers from 11 -26!
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    Sorry, in that case your question makes no sense whatsoever.
    Try to restate the question is a understandable form.
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    I'm sorry, maybe that "random" is confusing. Let's say I need to find out how many combinations of 10 numbers from 11 - 26 contain the numbers 11, 12, 13, 14!!
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    Quote Originally Posted by aimpro2000 View Post
    I'm sorry, maybe that "random" is confusing. Let's say I need to find out how many combinations of 10 numbers from 11 - 26 contain the numbers 11, 12, 13, 14!!
    The answer to that particular question is: \dbinom{12}{6}.
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    thanks
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    Quote Originally Posted by aimpro2000 View Post
    Given are all the possible 10 number combinations (NOT permutations) of the numbers 11 - 26 (8008 combinations), and 4 random numbers from the 11 - 26 sequence.
    How do I find out how many of the 8008 combinations contain
    a. all 4 random numbers?
    b. 3 of the 4, 2 of the 4, 1 of the 4 and 0 of the 4 random numbers respectively??
    (a)

    There are 16 numbers from 11 to 26 inclusive.
    If you set aside your four random numbers, you want to choose 6 of the remaining 12 to go with them.

    There are \displaystyle\binom{16-4}{10-4} ways to do that.

    (b)

    There are \displaystyle\binom{4}{3} ways to choose 3 of the 4 random numbers.

    For each of those choices there are \displaystyle\binom{16-4}{10-3} ways to choose 7 numbers to go with them,
    such that 3 of the 4 random numbers are chosen.

    There are \displaystyle\binom{4}{2} ways to choose 2 of the 4 random numbers.

    There are \displaystyle\binom{16-4}{10-2} ways to match them with 8 others that do not belong to the random numbers.

    Same strategy for the remaining selections.
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  9. #9
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    Quote Originally Posted by aimpro2000 View Post
    any 4 numbers from 11 -26!
    *ahem* 26! is quite a large number ....Perhaps you just mean 26.
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