90720 distinguishable permutations.
1) Considering that the word CLEVELAND, actually has 9 letters and there is the repetition of E, L.
2) And the question makes no comment about restraining repetitions we can write considering the repetition formula:
[tex]\frac{nPr}{x_1!x_2!}=\frac{9\times8\times7\times6\times5\times4\times3\times2\times1}{2!2!}=\frac{362880}{4}=90720[/tex]
In this formula above, we can place the repeated letters (2 letters E and L) as x_1 and x_2, and on the numerator the number of letters, in this case, 9.
3) So there are 90720 possibilities of distinguishable permutations
