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Suppose that I have 5 balls with different weights. A scale tells me which one ball is heavier than another. I have to write down the the pairs of balls I use before I use the scale. Is it possible to write down at most nine pair of balls which tells me the order of balls weights?

I guess the answer is no as 4+3+2+1=10>9 but I have no proof.

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As you point out, there are 10 pairs of balls. If we renumber the balls, any weighing of 9 pairs will omit one pair, so let us say we write down all the pairs except 4 vs 5. Any ordering of the weights that has 4 and 5 neighbors will be indistinguishable from the same ordering with 4 and 5 reversed. So you are right that if you have to specify the weighings in advance there are cases you cannot resolve with 9 weighings. You could tell 14253 from any other order, but could not tell 14523 from 15423.

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    @RossMillikan That's a good point - I'd missed that particular detail. At least in theory you might even be able to do 7 (since 120<128), but that seems unlikely. I'm sure TAOCP vol. 3 has something to say about this, but sadly my copy's at home right now.2012-08-28
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You must know the number of different pairs you can do with $5$ balls (the minimal number of checks) which is $\frac{5!}{2!.(5-2)!}=10$