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Guess who’s a winner?

February 1, 2009 am28 6:24 am

Monday Math Madness is an every other week contest, divided between two blogs: Wild About Math and blinkdagger.

Many of the contests (most?) are counting problems. My kind of stuff. And I enter about half the time. I like to solve the problems as I compose the e-mail, without, if I can help it, pencil or paper.

And last week, #24, I won.

And next week, #25, was fun. I solved it kind of tipsy last night. But I don’t think I am eligible anymore.

My prize: some sort of gift certificate, I will make it a prize for a puzzle solution in my combinatorics class. I might even use #25… Think a high school kid could do it?

And that’s the thing of it… these puzzles are hard, but… it’s what I teach my students. I assume that in a class of 25, if I had run the MMM contests every two weeks, figure that’s 8 contests, we would have had at least one correct solution, and perhaps a few.

But they are still fun for me, and I encourage you to take a look. And I figure that it’s ok to take the prize if I am using it to motivate kids.

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6 Comments leave one →
  1. February 1, 2009 pm28 3:20 pm 3:20 pm

    Aw man, I’m so jealous, I never win. Congrats!

    • February 1, 2009 pm28 9:18 pm 9:18 pm

      That’s what I used to say! Keep trying. And thanks for the congrats.

  2. Jewkulak permalink
    February 2, 2009 am28 2:58 am 2:58 am

    Congrats again! I had the same solution but I guess the random number generator was a little kinder to you. I’m in high school btw so it is definitely possible for HS students to solve these problems. Your solution to MMM #24 was a lot more concise than mine though. I had to use multinomial and some other stuff.

  3. February 2, 2009 am28 4:31 am 4:31 am

    Congrats on solving 24. My write up was not so nice (they published someone else’s), but my approach to 25, they told me, was creative, and while I can’t win again, I’m hoping they run that one.

    My solution to 24, ugly as it may have been:

    There are 5^8 different “sets” that can be played.

    Good mixes of the five songs could be:
    41111 Pick one song to be the “4”, (5 ways), then rearrange all (1680
    ways or 8!/[4!1!1!1!1!])
    32111 (20 like this) Pick one song to be the “3” and another to the
    the “2” (20 ways (5*4)), then rearrange all (3360 ways or
    8!/[3!2!1!1!1!])
    22211 (10 like this) Pick three songs to be doubled (10 ways, C(5,3))
    then rearrange all (5040 ways, or 8!/[2!2!2!1!1!])

    So (5*1680 + 20*3360 + 10*5040)/(5^8) = 32.256%

  4. February 2, 2009 am28 4:32 am 4:32 am

    Interesting, those 8!/(a!b!c!d!e!) do look like multinomial coefficients. I hadn’t made the connection.

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