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Sagot :
From the word EPIPHANY, we can see that:
E,I,H,A,N,Y are unique and P repeats twice. Then we would have a permutation with repetition. Let's state some data to solve this problem:
n=8 (number of letters)
Repetitions of the letter E: 2
Then:
[tex]\begin{gathered} P_{}(n;a,b,c\ldots)=\frac{n!}{a!b!c!\ldots}^{_{}}_{} \\ \Rightarrow P(8,2)=\frac{8!}{2!}=\frac{8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}{2\cdot1}=\frac{40320}{2}=20160 \\ \\ \end{gathered}[/tex]Therefore, we can make 20160 arrangements using the letters EPIPHANY
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