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Thread: Combinatorics

  1. #1
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    Combinatorics

    1. There are two given sets, the first one with n elements and the second one with m elements. In how many ways can you rearrange all the elements, so that the first p elements are from the first set, and the last q elements are from the second set?

    Is my solution correct?


    Combinatorics-zadacha7.jpg

    2. How many different n digit numbers can be made with the numbers 3 and 5?

    Is the solution solely 2^n, or 2^n - 2 (without counting the n digit 333...3 and 555...5)? The 3 AND 5 confuses me. Do both 3 and 5 have to appear in the number?

    Thanks in advance.
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  2. #2
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    Re: Combinatorics

    Quote Originally Posted by MaryLaine View Post
    1. There are two given sets, the first one with n elements and the second one with m elements. In how many ways can you rearrange all the elements, so that the first p elements are from the first set, and the last q elements are from the second set?
    I would make things clearer: $0<p\le n~,~0<q\le m,~\&~p+q\le n+m$.

    Answer: $\left( {{P_p}^n} \right)\left( {{P_q}^m} \right)$ where $\left( {{P_k}^n} \right)$ means a permutation of n items taken k at a time.

    Quote Originally Posted by MaryLaine View Post
    2. How many different n digit numbers can be made with the numbers 3 and 5?
    Is the solution solely 2^n, or 2^n - 2 (without counting the n digit 333...3 and 555...5)? The 3 AND 5 confuses me. Do both 3 and 5 have to appear in the number?
    I would answer it as $2^n-2$. But the question needs clarification.
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  3. #3
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    Re: Combinatorics

    Why the permutations but not the variations?
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