## anonymous one year ago Permutation: how many four letter word can you make out of SUMMER?

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1. Michele_Laino

numbers of permutations of n elements are: $\Large n!$

2. anonymous

would it be 5! * 2?

3. Michele_Laino

we have to count how many subset of four elements I can make with 6 elements

4. anonymous

How would I do that?

5. 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

6. Michele_Laino

$\Large {D_{n,k}}$ isa called the numbers of simple dispositions

7. anonymous

so is it 5 P 4 + 6 * 4 P 2?

8. Michele_Laino

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

9. Michele_Laino

$\large \begin{gathered} {D_{n,k}} = n \cdot \left( {n - 1} \right) \cdot \left( {n - 2} \right) \cdot \left( {n - 3} \right) \cdot ... \cdot \left( {n - k + 1} \right) = \hfill \\ = 6 \cdot 5 \cdot 4 \cdot 3 = ...? \hfill \\ \end{gathered}$

10. Michele_Laino

sorry for my first post, I have misunderstood your question!

11. anonymous

Ahah no worries :) tyyy

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