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Help me please I'm stuck unless problem in my practice

Help me please I'm stuck unless problem in my practice-example-1
User Sloosh
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1 Answer

5 votes

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:


(nPr)/(x_1!x_2!)=(9*8*7*6*5*4*3*2*1)/(2!2!)=(362880)/(4)=90720

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

User Jay Edwards
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