How Many Ways Can The Letters In Math Be Arranged Combinatorial Math Concepts Needed For Upcoming Article on Parallel Universes

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Combinatorial Math Concepts Needed For Upcoming Article on Parallel Universes

In an upcoming article, I intend to prove, using nothing more than high school math, that there is an exact replica of you in a parallel universe. In fact, in this parallel universe, there is another planet Earth with everything we have on our planet. I hope you look forward to this exciting science adventure. Before I do, though, it’s worth taking a moment to remind all of you readers of something you learned in high school math class.

If you remember any high school math, you’ll remember the following problem: In how many unique ways can the letters in the word MISSISSIPPI be arranged? Notice that there are repetitions of some letters: I and S appear four times each, while P appears twice.

Since it is an arrangement, the order matters, meaning that MISSISSIPPI is a different arrangement than IMSSISSIPPI, obtained by changing only the first two letters. If there were no repetition, we would use the permutation formula symbolized by 11P11, and find that there are almost 40 million arrangements (39,916,800 to be exact). Because of the repetition, many of these arrangements are the same, so we must divide this result by products of factorials for each of the repeating letters. (As a reminder, four factorials, symbolized by 4!, means 4 times three times two times one, which equals 24.) So four factorials equal 24, and there are two for the letters I and S. For the letter P, we use two factorials that are equal to two. So we need to divide the huge number above by the product of 24 times 24 times 2:

11P11/((4!)(4!)(2!)) = 39,916,800/((24)(24)(2)) = 34,650.

Without the repetition, of course, there are far fewer arrangements. That’s all you’ll see in most high school math books for permutations with repetition.

But what if one of your bright students asks the following question: How many unique arrangements can be made from the letters of the word MISSISSIPPI if you want to make arrangements of fewer than 11 letters? For example, how many unique arrangements of five letters can be formed? This new problem is not difficult, but it will be immeasurably helpful if we first go back to the original problem and look at it in a different way.

In the original problem, we wanted to form arrangements using all the letters of the word. Note that there are only four different types of letters in the word MISSISSIPPI: in order of decreasing frequency of occurrence, they are I, S, P, and M. Now we can begin the problem by asking: In how many ways can we arrange the four I’s in the 11 places we have to fill? Since the four I’s are indistinguishable, we would use the combination formula represented by 11C4 and get 330 ways. There are seven places left to fill, so we turn to the letter S and ask how many ways we can arrange the four S’s in those seven places: that would be 7C4, or 35 ways. There are three ways to arrange the two P’s in the three remaining points, which we get from 3C2, and finally 1C1 gives us a way to put the M in the last remaining place. The counting principle tells us to multiply these four numbers together to get the total number of ways these letters can be arranged:

(11C4)(7C4)(3C2)(1C1) = (330)(35)(3)(1) = 34,650.

Notice that we got the result we got above with a single permutation! It is worth noting that the counting principle gave us the unique number of arrangements (permutations) after using combinations to take care of all the repetition. Very nice, don’t you think? In certain situations, therefore, combinations + principle of counting = permutations.

Now, back to our brilliant student who has been patiently waiting for an answer. Armed with what we now know, it’s easy to answer your question. If we are forming arrangements of five letters, we start again and ask ourselves: In how many ways can the four I’s be arranged to occupy the five places? That would be 5C4, giving 5 ways. We have just filled four of the five spots, with only one left to fill. There are three types of letters left, so we can simply multiply by three and we have our answer:

(5C4)(3) = (5)(3) = 15.

This article will help all you teachers who are at the mercy of little combinatorial geniuses opening up in your classrooms. (For more, see Bettina and Thomas Richmond’s A Discrete Transition To Advanced Mathematics, published by the American Mathematical Society. In any case, you’re now ready to explore parallel universes, so stay tuned!

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