Saturday, April 30, 2005


In last Friday's Dilbert, Dogbert says that a customer has 25 possible solutions to try for his broken computer, but they must be tried in every possible combination. The customer replies that this results in "625 things I'd have to try". I don't get it. I'm no mathematician, but should that be 2 to the 25th, or 33,554,432? Where's 625 come from?


  1. Mathematically speaking, in combinations (like a lottery, but unlike lock combinations), the order does not matter. When Dogbert said "in every possible combination," I think he meant permutation, i.e. the order of the possible solutions was important.

    The formula for choosing k objects from a group of n objects is n!/k!(n-k)! where order is not important (combination), or n!/(n-k)! where order is important (permutation). There's only 1 combination of doing 25 things, but 25!, or 1.551121 × 10^25, different orders for those 25.

    If Dogbert has to do every combination, meaning trying each individually, followed by each pair, followed by each combination of three, then he would be having a similar experience to what I am dealing with with this @#$%@#$ Dell from one of my clients.

  2. You're right, Jonathan. The 625 did come from 25^2, but it's not the right answer. "Every possible combination" means that the first possible solution can be in or out of what you're trying, which gives you 2 things to try. Independently, the second thing could be in or out, which gives you (cumulatively) 2x2 things to try. Same with the third, which gives you 2x2x2, etc. So there are 2^25 (or maybe 2^25 - 1, if you can't do nothing). The interesting question is whether Scott Adams knew that he was giving the wrong answer.