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Pick up a random permutation in S5(assuming all elements have the equal chance to be picked). Find the probability that the sum of the first three entries of σ is less than or equal to sum of last two.

My try: I mean there will be 5! different combination possible, do I have to look at each of it?

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1 Answer 1

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Hint: the permutations you want are just : $ (...,5,4)$, $(...,5,3)$, $(...,4,5)$, $(...,3,5)$, $(...,2,5)$ and $(...,2,5)$ and you don't care about the order of the other elements where I put the dots. So the probability is: $$\frac{1}{5}\cdot\frac{1}{4} + \frac{1}{5}\cdot\frac{1}{4} + \frac{1}{5}\cdot\frac{1}{4} + \frac{1}{5}\cdot\frac{1}{4} + \frac{1}{5}\cdot\frac{1}{4} + \frac{1}{5}\cdot\frac{1}{4}.$$

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  • $\begingroup$ Won't we include (...,5,2) and (...,2,5) as it asks for less than or equal to? $\endgroup$
    – max
    Feb 10, 2016 at 23:34
  • $\begingroup$ So, there will be six cases like this ( 4 you have mentioned and 2 by me). And the probability should be (6* 3!)/120? Because we can arrange the first three elements in 3! ways. $\endgroup$
    – max
    Feb 10, 2016 at 23:36
  • $\begingroup$ You are right!! :D $\endgroup$
    – Onil90
    Feb 10, 2016 at 23:37
  • $\begingroup$ 4+2+1 =7 ? How come is this 8? $\endgroup$
    – max
    Feb 10, 2016 at 23:37
  • $\begingroup$ sorry! I'm super tired (and slightly drunk). $\endgroup$
    – Onil90
    Feb 10, 2016 at 23:39

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