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Sagot :
Answer:
[tex]1\, 260[/tex].
Step-by-step explanation:
There are [tex]n![/tex] ways to order [tex]n[/tex] distinct elements without replacement. However, in this question, there are two groups of elements that are not distinct. These duplicates need to be eliminated from the count.
For example, notice that if the two letters "[tex]\texttt{l}[/tex]" in the word are labelled [tex]\texttt{l}_{0}[/tex] and [tex]\texttt{l}_{1}[/tex], there would be [tex]2![/tex] ways to order them within every possible way to arrange the seven letters:
- [tex]\texttt{m i }\texttt{l}_{0}\texttt{ }\texttt{l}_{1}\texttt{ i o n}[/tex].
- [tex]\texttt{m i }\texttt{l}_{1}\texttt{ }\texttt{l}_{0}\texttt{ i o n}[/tex].
Similarly, if the two letters "[tex]\texttt{i}[/tex]" are labelled, there would be [tex]2![/tex] ways to order them within each possible spelling of the word.
Apply the following steps to find the number of distinguishable arrangements of the letters in this [tex]7[/tex]-letter word:
- Find the number of orderings as if all [tex]n = 7[/tex] letters are distinct.
- Divide the count from the previous step by [tex]2![/tex] to account for the letter [tex]\texttt{i}[/tex] , which was repeated for a total of two times.
- Similarly, divide the count by [tex]2![/tex] again to account for the letter [tex]\texttt{l}[/tex].
Hence, there would be a total of [tex]7! / (2! \times 2!) = 1\, 260[/tex] distinguishable arrangements of the letters.
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