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anonymous

  • one year ago

Permutation: how many four letter word can you make out of SUMMER?

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  1. Michele_Laino
    • one year ago
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    numbers of permutations of n elements are: \[\Large n!\]

  2. anonymous
    • one year ago
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    would it be 5! * 2?

  3. Michele_Laino
    • one year ago
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    we have to count how many subset of four elements I can make with 6 elements

  4. anonymous
    • one year ago
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    How would I do that?

  5. Michele_Laino
    • one year ago
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    more precisely, we have to count how many subsets of four elements, we can make with 6 elements, being those subset different by the order of their elements or by the type of elements. In general, if we want to get howm many k elements subsets we can form with n elements, we have to apply this formula: \[\large {D_{n,k}} = n \cdot \left( {n - 1} \right) \cdot \left( {n - 2} \right) \cdot \left( {n - 3} \right) \cdot ... \cdot \left( {n - k + 1} \right)\] Now, in your case k=4 and n=6

  6. Michele_Laino
    • one year ago
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    \[\Large {D_{n,k}}\] isa called the numbers of simple dispositions

  7. anonymous
    • one year ago
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    so is it 5 P 4 + 6 * 4 P 2?

  8. Michele_Laino
    • one year ago
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    if we apply my formula above, we get: \[\Large \begin{gathered} {D_{n,k}} = n \cdot \left( {n - 1} \right) \cdot \left( {n - 2} \right) \cdot \left( {n - 3} \right) \cdot ... \cdot \left( {n - k + 1} \right) = \hfill \\ = 6 \cdot 5 \cdot 4 \cdot 3 = ...? \hfill \\ \end{gathered} \]

  9. Michele_Laino
    • one year ago
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    \[\large \begin{gathered} {D_{n,k}} = n \cdot \left( {n - 1} \right) \cdot \left( {n - 2} \right) \cdot \left( {n - 3} \right) \cdot ... \cdot \left( {n - k + 1} \right) = \hfill \\ = 6 \cdot 5 \cdot 4 \cdot 3 = ...? \hfill \\ \end{gathered} \]

  10. Michele_Laino
    • one year ago
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    sorry for my first post, I have misunderstood your question!

  11. anonymous
    • one year ago
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    Ahah no worries :) tyyy

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