Question 1206532
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9 letters; of them "e" repeats twice and "n" repeats twice.


The number of distinguishable permutations (arrangements) is  

    {{{7!/(2!*2!)}}} = {{{(7*6*5*4*3*2*1)/4}}} = 7*6*5*3*2 = 1260.


The number of favorable arrangements is 1.


The probability is {{{1/1260}}}.
</pre>

Solved.


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The way on how the problem is worded in the post is a mathematical tragedy.


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