# Guess who’s a winner?

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.

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

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

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.

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%

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