A short while ago in in “Average, Percent, and other Misunderstood Math Terms”, I wrote about the fact that even simple math was particularly hard in application. I got three replies that suggested I had committed some unpardonable sin, and two even reminded me of the hassle when Mattel released the talking Teen-talk Barbie, who expressed a similar concern in her tiny little recorded voice by adding “Math is Hard” to her other stereotypical exaggerations of modern teens, such as “I like to shop.” I figured if a merchandising mishap 17 years ago by a toy company is the best they can do to challenge my statement, I MAY actually be right.

But to be fair, I wanted to explain why I thought math was hard, and while I was thinking about it, a couple of VERY good mathematicians brought up a minor variation of a somewhat easy and common high-school text book problem.. Only this problem is VERY hard. Its a high school problem that IS in a high school textbook, although it probably shouldn’t be, and it had the wrong answer in the back of the book .. *we know it is wrong, …too big.. but we don’t have the right answer yet*… keep in mind that teen talk Barbie is older than most of the students in my classes who would have been expected to solve the problem. You can still buy her on the Amazon link in the last sentence, and she still says “Math is hard”,… hang in there Barbie.

So here is how an easy problem crosses the border from easy to incredibly hard in a few steps. We begin with a linear permutation. To make the problem even easier than the one in the textbook, lets deal with only five beads.

Question one, easiest… Five different colored beads are to be placed in a row. How many ways can it be done…Easy, five factorial or 120 ways. You have five choices for the first color, then four choices for the next .. etc.. so the answer is 5(4)(3)(2)(1) = 120…

Question two, a minor variation… How many ways can the five colors be laid in a circle? Now we have a clinker (think Christmas Story and the furnace). Red, orange, yellow, green, blue is the same as orange, yellow, green, blue, red. It is also the same if we write them starting with yellow, green… etc.. so in the case of a circle, we can only get 5 factorial divided by 5 or 24 different circles.

Question three, a spatial flip.. Now instead of putting them in a circle on the ground, we put them on a hoop of wire, for a necklace or bracelet or something. Ok, still a circle so 24,… but wait.. we can reverse the necklace so that now blue, green, yellow, orange, red is the same as red, orange, yellow, green, blue.. and similarly for any list of colors, so we have to divide by 24 by two and now we are down to twelve unique necklaces that can be made… think goodness it can’t be any harder than that….(gotcha’)

Question four (really hard and if you go up to 20 beads, as the text book did, a total mess) What if there are not five colors, let’s be really easy, what if there are three red and two blue? Now the problem requires us to look at every possible order of the beads and decide how many ways it can be reflected in space or rotated around the wire… You could have 3 reds together and 2 blue…or you could have two reds, a blue, a red, and another blue… We could also have red, blue, red, blue, red… is that it? What if we had a blue, two reds, a blue, and another red… NOPE, that one is already counted.. can there only be three ways to order these five beads on a necklace. Math got hard because all the way down to the last problem we had some formula approach we could use if we understood the problem… and then… it all fell away..

If you need a challenge, the original problem had fourteen beads of one color and six of the other.. try your luck.

A very similar problem uses the letters in the spelling of Mississippi.. Writing words is a linear permutation, so we don’t have to worry about flipping the necklace or circular similarities, but that doesn’t make it easy. Students are taught, using the same rule we used for the beads in a row, that with six distinct letters there are 6 factorial orders (that’s 720) . With 11 letters there are 11 factorial (a lot more than 720, in fact almost forty million) . But then if we repeat letters, we show them that they have to divide by how many letters appear more than once..

If you write out all the 11 factorial arrangements of the eleven letters, all of them have two p’s that are indistinguishable. If we switched the two p’s no one would know; so we have to divide 11 factorial by 2. But then, there are four letters s in the word, so you have to divide by the 24 ways they could be switched around, and then there are four letters i, so we have to divide by 24 yet again. Finally we have reduced the unique orderings of Mississippi down to only 34,650 (IF I did that right) . You can list them out to check, but that may take a while… If you want to try an easier one that you can see how, try finding all the possible orders of TOOTH; there should be only thirty…

Now we make a subtle change in the problem and make it horrible, what if we don’t use all the letters? What if we only use eight of the letters in Mississippi at a time? One word has all four of the letters s, another might only have three, and some would only have one s (with eight letters, there has to be at least one of the letters s in the word, can you see why?) Try your luck figuring out how many three letter “words” you can make with tooth.

I think that is the big difference between math and other subjects, that razor fine line where easy problems become totally intractable. Maybe math is easy, and I (and Barbie) are just not too bright). By the way, how many eight letter "words" could YOU find using the letters in MISSISSIPPI?