I want to tell you a story. Everyone likes a story, right? But at the heart of my story lies a confession that I am afraid will shock many readers. People assume that because I teach math, blog about math, give advice about math on internet forums, and present workshops about teaching math — because I do all this, I must be good at math.

Apply logic to that statement. The conclusion simply isn’t valid. …

**Update:** This post has moved.

Oh cool. I can hang out with you then.

I’m finding the directive to, “Concentrate and try harder” from those who know more math than I do isn’t very helpful.

I figured out the dealio with summing an infinite geometric sequence. It was the first time in my life I’ve understood a “limit.” It’s a great feeling. The best analogy that I can come up with is learning to play the piano. I have no talent and the songs are old, but they are new to me when I learn to play them for the first time. It’s worth doing when no one else knows.

There is always somebody better, but there’s many people who don’t play at all.

My goal in math is to work my way through an Abstract Algebra book. I have a particular one in mind and I want to do it because at the very end of the book is a proof that PI is irrational and it’s just killing me to not know it. First, I have to survive the set identity proofs 😉

Wonderful 🙂

I am happy to have contributed so to your adventures in mathematics.

By the way, the Riemenn hypothesis is VERY nontrivial and difficult to understand. Certainly not a an easy place to venture into higher math. After all, it still stumps all the great mathematicians since!

By the way, the professor is still waiting for YOU to offer a probability problem that stumps you. The relevance is always higher when it is a problem you want to solve 🙂

Oh, so

that’swhy you haven’t come back with something for me to test myself on? Unfortunately, I haven’t found another problem that stumps me—MathCounts is such a jumbled hodge-podge that it’s not a good resource for any particular type of problem. And I know that if I try to make something up for myself, it will be too obviously a copy of the problem we just worked through. Since you enjoy combinatorics, I was hoping you might be able to offer something that was similar-yet-different enough to be a challenge to me.About a year ago I read Marcus du Sautoy’s “Music of the Primes”, wonderfully written history of the Riemann Hypothesis, as well as a history of number theory.

It’s been fun to find out as an adult that there is more to math than calculating how quick swimming pools fill up with pipes of three different rates or how far away a cannonball will land.

Unfortunately, I don’t have my books with me here in Israel, so I don’t have some immediate “pull out” things. But I’ll do some search for you 🙂

And Mytle, it’s a lovely book! Written very well as such books go.

Just a side note: I am working on Primes as I found some quite interesting things about their structure. I’m trying to find more before I publish them…in a mathematics journal actually, but since I do it on and off when I have some free time, it might take a while.

Is the answer (6^2 * 5^2)/6^4 = 25/36 ?

No, that fraction is much greater than the actual answer. I will include the answer at the bottom of this comment, so if you are still trying to work the problem on your own, you may want to stop scrolling down.

This weekend, I was working through last year’s state MathCounts competition in preparation for our class this week, and I ran into a really nasty combinatorics problem. And I was able to solve it. Hooray! Of course, I also ran into a few that I missed—figuring out how to count all the possibilities can be a challenge, and I was trying to work quickly, to simulate the stress my students would experience in taking the test—but being able to solve the hardest one correctly was a great encouragement.

Anyway, the answer to this problem is thirty-five seventy-secondths. (Hmm… Is that how the fraction would be written? My kids love to call fractions like this

seventy-twoths, rhyming with “tooth.”)Myrtle and Moti, thank you for the book recommendation. I got

The Music of the Primesfrom the library, and I am thoroughly enjoying it.thanks for bringing my attention to this

(in a recent post); very well-said.

i seem to’ve worked out the wrong answer here …

getting back to work …

Yeah, me three! I first interpretted it to mean that no two people sitting next to each other get the same as *each other* where everyone rolls their die exactly once. I get

(6*5*5*4)/(6*6*6*6) = 25/54

Or, perhaps the interpretation was supposed to be that everyone rolls their die, and then they roll it again to see if they get the same thing they got the first time. And, no two sitting next to each other do get the same thing they got on their first roll (as opposed to what the guy sitting next to them got). On this interpretation, I do more probability than combinatorics. The probability that one person rolls the same thing twice is 1/6 and you have 4 independent experiments taking place. (One person rolling the same thing twice does not influence the likelihood of the next person rolling the same thing twice.) So, you really just have to add up the probabilities, then, of the various outcomes that include no two people sitting next to each other rolling the same thing twice. Either no one rolls the same thing twice — (5/6)^4, one person rolls the same thing twice and no one else does — 4*(1/6)(5/6)^3, or two people sitting across from each other roll the same thing twice and the other two don’t — 2*[(1/6)^2][(5/6)^2]. Since these are all mutually exclsive events,

(5^4+4*5^3+2*5^2)/6^4 =

(5^2+4*5+2)5^2/6^4 =

(47*5*5)/(6*6*6*6)

It turns out that 47 is prime. So this just comes out to

1175/1296

I cannot figure out how you get (in any simple combinatorial way) 35/72 = (7*5)/(6*6*2) = (7*5*3*6)/6^4. How do you get 7*6*5*3 as the number of ways to do something, here? Is it like 6*6*5*3+6*5*3? Or, maybe 6*5*4*3+6*5*3*3? I can’t figure out what you are counting up, here. 6^4 is the total number of possible outcomes of all four rolling their die where order matters. (The people are ordered 1 to 4 so that 6,5,4,3 is a distinct outcome from 3,4,5,6, for instance). 6*5*4*3 is the number where they each roll something different (again, where order matters).

…?…

(Combinatorics is always a humbling experience…. Perhaps Vlorbik will post the complete solution or Denise if she is still reading these comments.)

Your first interpretation is correct, but you have not counted all the possibilities. The difficulty comes because the table is round, so that person #4 is next to both #1 and #3. So you have to look at two separate cases: these guys roll different numbers, or they roll the same number. Only then can you tell what the options are for #4.

Same number: 6*5*1*5 cases.

Different number: 6*5*4*4 cases.

OIC. I have 6*5*5*4. On the one hand, the second factor of 5 includes the possibility that the first guy and the third guy roll the same thing. But, the 4 assumes that they definitely rolled different things which undercounts the fourth guys possibilities when one and three do roll the same thing.

Very clever….

Aside from your good post about math problems and puzzles you are also good in telling story. Base on your story, I should never be afraid of math….

Enjoy solving,

John

Here’s the approach I used. Instead of trying to directly compute the probability of no two people sitting next to each other, compute the probability of two people sitting next to each other getting the same number. So first count all the different possible outcomes, call that number A, then count all the different ways in which two people sitting next to one another get the same number, call it B, then your answer is 1 – (A/B).

Oh, no!!!! I’ve found another place online that will consume me and eat away at my “free time” (What’s that???)

This looks like a really interesting blog and I will be back (sigh) to look at it some more.

I am also a math teacher. I also love math. It can be all consuming, can’t it?

I can totally identify with your description of how the light turned on about the combinatorics problem while you slept. Been there, done that. : )

I will go to bed now and think about your combinatorics problem- maybe I’ll come up with the solution.

Trouble with a probability problem is no indication of not being good at math – they’re hard to analyze properly. I’m sure you’ve read the hoopla about the 3 doors problems (aka the Monty Hall problem). When Marilyn vos Savant wrote about it in Parade,

mathematicianstold her she was wrong. (She wasn’t.)Sounds to me like you’re good at the math you’ve dedicated your attention to.

very honest Denise. I am not that good at math either.