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December 5, 2008 pm31 4:59 pm

Miss Cornelius sent me to play a little “quiz” – Could you pass high school?

Now, sounds like fun, but also sounds like they didn’t really have it right – pass math, graduate high school? (Or is it, be graduated from high school?)

Anyhow, and sure enough, 10 light-weightish questions, but take a look at this one:

Jennifer has 2 skirts, 3 tops, and 3 pairs of shoes. How many outfits can she make?

Their choices are 8, 12, 18, and 24. I know they wanted 18. But how am I getting 119?

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8 Comments leave one →
  1. David Petersen permalink
    December 5, 2008 pm31 5:11 pm 5:11 pm

    Well, if she lives in Tennessee like I do, “no shoes” could be an option and 24 would be right. I suppose “no top” or “no skirt” would also be options, but inappropriate. Other options are to mismatch shoes or wear multiple skirts/tops at the same time (as some girls like to do).

  2. December 5, 2008 pm31 7:34 pm 7:34 pm

    Well, even if you added barefoot, topless, and bottomless (scandalous!) as options that would only get you to 3*4*4 or 48. Do you have her wearing shoes on her hands and skirts on her head or something?

  3. David Petersen permalink
    December 5, 2008 pm31 7:41 pm 7:41 pm

    Well, if you count “left shoe” and “right shoe” separate and allow for nudity and barefeet, you get 3*4*4*4 = 192 and the other options of “tops” being sweater, t-shirt, and spaghetti-straps (so that some girls might wear 2 or 3 at a time) there are more options than that. I don’t quite see the 119 combo that you got

    • December 5, 2008 pm31 8:01 pm 8:01 pm

      Well, you hit almost everything I thought of. I think there are just two more details…

      Getting to 119 from 192…

  4. December 5, 2008 pm31 8:56 pm 8:56 pm

    Naively, there are \sum_{k=1}^n k! * (n k) ways to wear n tops, where (n k) denotes a binomial coefficient. (You can choose k of the n tops, and the order matters.) For n=3 that’s (3 1) + 2 * (3 2) + 6 = 15. In reality, however, you’d want to put some sort of preordering on the tops according to the “covers up” relation, and only allow orderings of the tops where none are covered by any further along in the ordering. If you wear a tank top under a sweater no one can see it and you might as well not be wearing it. But for now let’s assume that none of the tops cover up any of the others. Skirts are similar; there are 2 + 2 = 4 ways to choose skirts.

    With shoes, you can choose one of three shoes for the left foot, and independently, one of three shoes for the right, for a total of 9 possibilities.

    Altogether, we get 15 * 4 * 9 = 540 outfits, or 15 * 4 * 16 = 960 if we allow bare feet, or 16*5*16 = 1280 if we also allow nudity.

    Of course, it is actually a trick question, because not all of these combinations count as “outfits”, the definition of which depends on what the clothing items actually look like and the assumed cultural context.

  5. David Petersen permalink
    December 5, 2008 pm31 10:05 pm 10:05 pm

    Another thought… (just to be snarky about it) The question is really, “How many outfits can she make?” and if she’s a seamstress (or cobbler), maybe she can make a lot. Either out of the material from skirts and top (or shoes) or from other material not mentioned.

    Here again the definition of “outfit” could be pushed to its limits and you’d also have to ask how old she is (i.e. how much time does she have) and all sorts of other variables could come into play, but the number could presumably be much much higher than situations we’ve been proposing here.

  6. December 8, 2008 am31 2:12 am 2:12 am

    Well,

    you really hit all the points. I allowed mismatched shoes, but not one shoe on, one shoe off. And I subtracted 1 – wearing absolutely nothing doesn’t meet my definition of ‘outfit’.

  7. Eric Jablow permalink
    December 23, 2008 am31 9:38 am 9:38 am

    I hate say it, but this discussion reminds me of the classic condom problem. I know, the page redirects to “glove problem”, but we all know the original version.

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