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Sagot :
This is an example of Permutation with Repetition. Notice that while we are arranging 8 letters in total, the letters D and E were repeated, which makes a big difference the way the permutation is calculated.
To illustrate, imagine one E being in bold (E) and the other being italicized (E). Two possible permutations would have been:
TEDEDUCA and
TEDEDUCA
They look different because of the bold and italics but since we don't have those in reality, the two permutations will simply be TEDEDUCA.
To solve this, we use:
[tex] \frac{8!}{2!2!1!1!1!1!} [/tex]
[tex]8![/tex] because we have 8 letters being arranged
the two [tex]2![/tex]'s represent the two letters, E and D, that were repeated twice
the four [tex]1![/tex]'s represent the other four letters that appeared only once.
Therefore, the number of distinguishable permutations of the letters of the word EDUCATED is
[tex] \frac{8!}{2!2!1!1!1!1!}=\frac{8!}{4}=10,080[/tex].
The answer is B.
For more examples of Permutation with Repetition, check https://brainly.ph/question/260845
To illustrate, imagine one E being in bold (E) and the other being italicized (E). Two possible permutations would have been:
TEDEDUCA and
TEDEDUCA
They look different because of the bold and italics but since we don't have those in reality, the two permutations will simply be TEDEDUCA.
To solve this, we use:
[tex] \frac{8!}{2!2!1!1!1!1!} [/tex]
[tex]8![/tex] because we have 8 letters being arranged
the two [tex]2![/tex]'s represent the two letters, E and D, that were repeated twice
the four [tex]1![/tex]'s represent the other four letters that appeared only once.
Therefore, the number of distinguishable permutations of the letters of the word EDUCATED is
[tex] \frac{8!}{2!2!1!1!1!1!}=\frac{8!}{4}=10,080[/tex].
The answer is B.
For more examples of Permutation with Repetition, check https://brainly.ph/question/260845
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