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# Combinatorics problem watch

1. The numbers 1,2,...,n for n>=4 are randomly arranged in a row.
What is the probability that the number 1 is somewhere to the left of the number 2?

(A) 1/2
(B) 1/n
(C) 1/(2(n-2)!)
(D) 1/(2(n-1)!)
2. Looking at all combinations (which are equally likely), you can swap 1 and 2 and it won't matter: you'll get the same combinations rearranged. Therefore there are an equal number of cases where 1 is before 2 and 2 is before 1, so the answer is A.
3. Thanks.
4. that was quick ...

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Updated: February 20, 2016
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