anonymous
  • anonymous
Permutation: how many four letter word can you make out of SUMMER?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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Michele_Laino
  • Michele_Laino
numbers of permutations of n elements are: \[\Large n!\]
anonymous
  • anonymous
would it be 5! * 2?
Michele_Laino
  • Michele_Laino
we have to count how many subset of four elements I can make with 6 elements

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anonymous
  • anonymous
How would I do that?
Michele_Laino
  • Michele_Laino
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
Michele_Laino
  • Michele_Laino
\[\Large {D_{n,k}}\] isa called the numbers of simple dispositions
anonymous
  • anonymous
so is it 5 P 4 + 6 * 4 P 2?
Michele_Laino
  • Michele_Laino
if we apply my formula above, we get: \[\Large \begin{gathered} {D_{n,k}} = n \cdot \left( {n - 1} \right) \cdot \left( {n - 2} \right) \cdot \left( {n - 3} \right) \cdot ... \cdot \left( {n - k + 1} \right) = \hfill \\ = 6 \cdot 5 \cdot 4 \cdot 3 = ...? \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
\[\large \begin{gathered} {D_{n,k}} = n \cdot \left( {n - 1} \right) \cdot \left( {n - 2} \right) \cdot \left( {n - 3} \right) \cdot ... \cdot \left( {n - k + 1} \right) = \hfill \\ = 6 \cdot 5 \cdot 4 \cdot 3 = ...? \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
sorry for my first post, I have misunderstood your question!
anonymous
  • anonymous
Ahah no worries :) tyyy

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