# What Does The Product Mean In Math latest 2023

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## Combinatorial Math Concepts Needed For Upcoming Article on Parallel Universes

In a future 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 enjoy this exciting scientific adventure. Before you do that, though, it’s worth taking the time to remind each reader of something you learned in high school math class.

If you remember a high school math lesson, you’ll remember the following problem: 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 it is an arrangement, the order matters, i.e. MISSISSIPPI is a different arrangement from IMSSISSIPPI, obtained by interchanging only the first two letters. If there were no repetition, we would use the permutation formula symbolized by 11P11 and find that there are nearly 40 million arrangements (39,916,800 to be exact). Due to repetition, many of these arrangements are the same, so we need to 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 which 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 vastly fewer arrangements. That’s all you’ll see in most high school math books when it comes to permutations with repetition.

However, what if one of your brilliant students asks the following question: How many unique arrangements can be formed from the letters of the word MISSISSIPPI if you wish to form arrangements of less than 11 letters? For example, how many unique five-letter arrangements can be formed? This new problem is not difficult, but it will be extremely 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 form arrangements using all the letters of the word. Consider that there are only four different types of letters in the word MISSISSIPPI – in descending order of frequency of occurrence, these are I, S, P and M. We can now begin the problem by asking: how many ways can we organize the four I’s into the 11 places we need 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 let’s move on to the letter S and ask how many ways can we arrange the four S’s into those seven places – that would be 7C4, or 35 ways. There are three ways to arrange the two Ps in the three remaining slots, which we get from 3C2, and finally 1C1 gives us a way to put the M in the last remaining slot. 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) = 34650.

Notice that we got the result we got earlier using just one permutation! Interestingly, the counting principle gave us the unique number of arrangements (permutations) after using combinations to support all repetitions. Very nice, don’t you think? In some situations, then, combinations + 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. If we form arrangements of five letters, we start over and ask: How many ways can the four I’s be arranged to fill the five places? That would be 5C4, giving 5 ways. We have just filled four of the five places, there is 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 teachers who find themselves at the mercy of little combinatorial geniuses making their way into your 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, you are now ready to discover parallel universes, so stay tuned!

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