### Instruction: The letters of the word `SIXTEEN' are randomly arranged.  What is the probability that the two E's are not next to each other?### Response: I can start by finding the total number of ways to arrange the seven letters, regardless of the position of the E's.
Since there are seven letters, but two are repeated, I can use the formula for permutations of a multiset: n! / (n1! n2! ...
nk!), where n is the total number of elements and ni is the number of times the ith element appears.
In this case, n = 7, n1 = 2 (for the E's), and n2 = n3 = ... = n6 = 1 (for the other letters).
So the total number of arrangements is 7! / (2! 1! 1! 1! 1! 1!) = 2520.
Next, I want to find the number of arrangements where the two E's are next to each other.
I can think of the two E's as a single unit, and then arrange the six units (EE, S, I, X, T, N) in any order.
The number of ways to do this is 6! / (1! 1! 1! 1! 1! 1!) = 720.
Therefore, the number of arrangements where the two E's are not next to each other is the difference between the total and the unfavorable cases: 2520 - 720 = 1800.
Finally, the probability of this event is the ratio of the favorable to the total outcomes: 1800 / 2520 = 0.7143.