Found Mixed-Results In How Teachers Should Respond To Scientific Questions 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 a future article, I plan to prove, using high school math, that there is an exact copy of you in a parallel universe. In fact, in that parallel universe, there is another planet Earth with everything that is on it that is on our planet. I hope you look forward to this exciting science adventure. Before I do that, though, it’s worth taking a moment to remind every reader about something you learned in high school math class.

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

Since this is an arrangement, the order is important, which is to say that MISSISSIPPI is a different arrangement from IMSSISSIPPI, obtained by moving only the first two letters. Without repetition, we 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 those arrangements are the same, so we need to divide the resulting product into 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 letter I and S. For the letter P, we use two factorials equal to two. So, we need to divide the big number above by the product of 24 times 24 times 2:

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

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

What happens when one of your bright students asks the following question: How many unique arrangements can be formed from the letters of the word MISSISSIPPI if you want to make arrangements with less than 11 letters? height? For example, how many unique five-letter arrangements can be formed? This new problem is not difficult, but it can be immeasurably useful if we first go back to the original problem and look at it in a different way.

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

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

Notice that we got the result we got earlier using a permutation! It is worth noting that the principle of counting gives us the unique number of arrangements (permutations) after we use combinations to take care of all repetitions. Very nice, isn’t it? In some situations, then, combinations + the counting principle = permutations.

Now, back to our brilliant student who is patiently waiting for an answer. Armed with what we know now, it’s easy to answer his question. When we make five-letter arrangements, we start over and ask: How many ways can the four I’s be arranged to fill the five spaces? This would be 5C4, giving 5 ways. We have just filled four out of five spots, only one left to fill. There are three remaining types of letters, so we can simply multiply by three and there is the answer:

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

This article will help all the teachers out there who find themselves at the mercy of the little combinatorial geniuses that happen to be working in their classrooms. (For more information, see A Discrete Transition To Advanced Mathematics by Bettina and Thomas Richmond, published by the American Mathematical Society. In any case, now you’re ready to explore the parallel universe, so stay tuned !

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