“Monday Math Madness” link

math puzzle
Photo by Behdad Esfahbod.

I have been busy with the end of Math Olympiad season and getting ready for the MathCounts state test this weekend, but I wanted to post this link before it’s too late. You have until Sunday evening to send in your answer to the first…

Monday Math Madness puzzle

Combinatorics is my weak area, but I gave it a try. How about you?

Edited to Add

Since the Monday Math Madness (link is to this week’s answers) will be an ongoing contest, I am adding it to the Blog Parties for Teachers widget in my sidebar. [The Monday Math Madness contest has gone out of business. I’m sure they wouldn’t mind, however, if someone else wanted to start it back up. Any takers?]

I will try to get each week’s post linked as soon as possible. With my busy schedule, I often run a day or two behind in updating the widget. As with the blog carnivals, I welcome email reminders from the puzzle contest hosts!


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12 thoughts on ““Monday Math Madness” link

  1. Efrique, my answer seemed relatively simple, too. But I always feel like I’m missing something on any combinatorics question. I just wish I knew enough to figure out what I am missing.

    Jonathan, I think he just wants any entries. Besides, I am still a struggling student when it comes to probability problems, so I figure I qualify either way. How do you make it cute?

  2. Without answering the question, I see at least 4 approaches:

    Find the number of selections that meet the criteria, and divide by the number of selections.

    Find the number of selections that don’t meet the criteria, and divide by the number of selections. Subtract from one.

    Find the probability of only 1 flavor. Then find the probability of exactly two flavors (do it once, and triple). Subtract these from 1.

    Pretend order matters. Find all the different “routes” to get to 5 via at least one of each. Decide that each is equally likely, find all the probabilities, and multiply by the number of routes (I did this, and sent it in).

    Pretend order matters. Find the probability of each dead end. Subtract the sum from one.

    I think the one I chose is a little cute. It looks massive, then collapses into something small.

    Jonathan

  3. A big part of my problem is that I am never quite sure what matters and what doesn’t. In this case, just because each post was assigned to a category at random does not mean that I have an equal chance of reading each category at random. Any blog has a finite number of posts, which means there is a finite chance that one category may be much smaller than the others, or even empty. But I don’t know enough about combinatorics to figure out how to count that factor — at least, not without more information.

  4. The answer is posted. And my answer was wrong; something that I thought shouldn’t matter (the order in which the posts are chosen) actually makes a big difference. As I said, this is my weak area. I can see in retrospect why my answer doesn’t work, but I will probably make the same mistake again the next time. I should stick to the simpler MathCounts-level problems until I master them.

  5. Mine was ok, but looked primitive compared to the chosen method. But then, I have to say, a lot of what I do looks primitive. (3n^3)^2? (3nnn)(3nnn). So I go for primitive.

    Try a simpler problem. How about 3 types, but only 3 books?

    aaa . aab . aac
    aba . abb . abc
    aca . acb . acc
    baa . bab . bac
    bba . bbb . bbc
    bca . bcb . bcc
    caa . cab . cac
    cba . cbb . cbc
    cca . ccb . ccc

    Can you find the 6 good choices? Out of 27?

    The 27 is 3x3x3
    The 6? The number of ways to rearrange abc.

    If you’d like, I’ll help extend this to bigger numbers.

    Jonathan

  6. Your list shows exactly why my approach didn’t work. The order in which the reader picks the posts matters, because it makes some combinations easier to get than others. There is only one way to choose 3 a‘s, but there are three different ways to get 2 a‘s and 1 b.

  7. Denise and others,

    I’m sorry that I missed this whole conversation. Between some paid writing projects and life I’ve not been tending to WAM! or to the Math blog community as much as I’d like.

    I’m delighted that there is discussion over this problem – it’s a fun one and tricky as well in terms of knowing how to count things.

    The next problem at Blinkdagger will probably have more of a logic bent to it. It’s not been decided for sure what their next problem is but the Blinkdagger guys like logic problems.

    My next problem, the first Monday of April, won’t be a combinatorics one. It’ll be a simple and elegant one to state but one that I’ve not seen an elegant solution to. Fortunately, elegance is not a requirement for winning!

    Thanks, Denise, for plugging the contest.

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