Wednesday, 27 November 2019

I'm sure I'm not that clever so ??????

I recently came across this problem in the first Irish Mathematics Olympiad. I thought I figured it out mentally, but then realized, if it is an olympiad problem, it is probably not that transparent. Soo....... I'm posting it and going to think more deeply about my attempt, and the solution, but encourage you to comment with your solution, and I will make sure all of them are added to the post. And the question is.... Drum Roll Please:
5. A person has seven friends and invites a different subset of three friends to dinner every night for
one week (seven days). In how many ways can this be done so that all friends are invited at least
once?

Your answer goes here........
at

3 comments:

  1. Joshua Zucker28 November 2019 at 18:11

    Seems like an inclusion-exclusion process might be needed?

    That is, count all the ways, subtract the ways in which 1 friend is missing, add back in the ways in which 2 friends are missing because those were double-subtracted the first time, and so on.

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  2. Pat's Blog2 December 2019 at 00:45

    Joshua,
    That was my approach, but from some other things I've been reading, they all seem to imply that every permutation of the orders is a "different way", So there are 7! times as many solutions as I had figured. I guess that's something olympiad contestants and number theorists pick up along the way. I can see the elements of my limited solution in theirs, but my numbers were Way smaller.
    Thanks for the input.
    Hope you are well
    Pat

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  3. Joshua Zucker8 December 2019 at 04:55

    I agree with them about the 7! because inviting people to dinner on Monday is not the same as inviting them on Wednesday.

    It still seems pretty straightforward for an Olympiad problem. In the US nowadays I would rate this as a viable AIME problem (relatively on the hard side, probably) but not hard enough except mayyyybe for #1 on the USAMO.

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