How Many Distinguishable Permutations Are Review With E as the Third Letter
Intro to Combinatorics
The Mississippi Counting Issues
The Multinomial Theorem Explained
Today I'one thousand continuing to talk about the fundamentals of combinatorics with the Multinomial Theorem, and what better way to do this than to tackle some archetype combinatorics issues 😉
Have you lot got the chops to solve these problems?
The Mississippi Counting Problems
Problem 1
How many total arrangements of the letters in MISSISSIPPI are in that location?
Problem 2
How many arrangements of the letters in MISSISSIPPI exist such that the 2 P's are separated?
Trouble iii
How many arrangements of the letters in MISSISSIPPI have at to the lowest degree 2 adjacent Southward's?
Notation:
If yous need a review of the basics, check out this intro to combinatorics post:
Solution #i: Permutations of MISSISSIPPI
Getting Started
In the terminal post nosotros discovered that we can find the number of unique permutations by using the Central Theorem of Counting.
Since MISSISSIPPI has 11 messages, draw eleven lines and fill each in with the number of available letter of the alphabet choices, e.g. xi options for the first, 10 for the second, and then on…
This is equal to 11! or 39,916,800 permutations.
But is this right for the unique permutations of the letters in MISSISSIPPI?
Consider this…
What happens if I switch the 3rd and quaternary messages in MISSISSIPPI and go out all else the aforementioned? Are those dissimilar permutations?
Nope.
The 3rd and fourth letters are both S's so switching them changes zippo. The above adding is an over count because of these repeat letters.
So how do we adjust for these extra repeated arrangements?
Adjusting for Repeats
To adjust our calculation nosotros demand to split up out the duplicate permutations. Finding them isn't too difficult.
First determine what causes repeated arrangements. This comes from letters that occur more than than once. For MISSISSIPPI that includes 2 P's, 4 I's, and 4 S'due south.
Let's showtime with the P'south. For every permutation, nosotros tin can brand an identical permutation with the P's in opposite positions. So to adjust for these duplications, we must separate past 2! (the number of ways we can arrange the 2 P's).
To adjust for the repeated I's, separate past the number of ways nosotros can adjust 4 I'southward, which is 4!.
Lastly to business relationship for the 4 S's divide by another four!.
There we become! There are 34,650 permutations of the give-and-take MISSISSIPPI.
The Multinomial Theorem
This process of dividing out repeated permutations is exactly what the Multinomial Theorem is!
The Multinomial Theorem says in lodge to count the number of distinct ways a gear up of elements with duplicate items can exist ordered all you demand to do is split the total number of permutations by the factorial of the quantity of each duplicate (turns out doing the math is easier than writing that judgement 😳 ).
So since MISSISSIPPI has 11 total messages with 2 P's, 4 I's, and 4 S's our adding is: 11!/(ii!four!4!).
Solution #2: No Adjacent P's
To solve this trouble we take to get a footling artistic. We need to count the ways we tin can make permutations so that no P's are side by side. Let'south start elementary and remove the P'due south altogether.
This of course is only ane arrangement of MISSISSII, how many means could we rearrange those messages? Using the Multinomial Theorem there are 9!/(four!iv!) or 630 permutations.
Okay, now let'southward deal with the P's. I'm going to draw in blanks before and after every alphabetic character of MISSISSII. These will represent the possible places we tin can insert P'southward, and because each infinite is separated by some non-P letter of the alphabet, we know these are all safe possibilities.
We have ten blanks and nosotros can choose any 2 blanks to insert P's at, so we demand to calculate ten choose 2 = 45 ways (click hither for recap on the "choose" combinations formula).
Alright, last step.
We calculated that there are 630 ways of rearranging the non-P letters and 45 ways of inserting P's, so to discover the full number of desired permutations use the bones principle of counting, i.e. multiply the values together.
Then there are 28,350 permutations of MISSISSIPPI where the P'southward are not adjacent.
Solution #3: At Least two Adjacent S'southward
Where to begin? Yous know, we could come at this problem from the contrary angle. Instead of counting how many permutations have 2 or more than adjacent S's, nosotros could discover how many permutations have no side by side South'due south and and so decrease that number from the total permutations of MISSISSIPPI.
Shall we effort that?
Okay, this is like to our last problem then. Start by removing the Southward's and finding the permutations of the remaining letters using the Multinomial Theorem.
We have 7 letters with 4 I'due south and 2 P's, so that's a full of 105 permutations.
Adjacent add blanks before and later on each letter to stand for places nosotros can insert Due south'south.
There are 8 blanks and we accept iv Southward's that can exist inserted into whatever of those blanks. So let's calculate 8 choose four:
And then multiply the results together.
That gives us the number of permutations of MISSISSIPPI without adjacent S's. Remember we want to find the reverse. We want to know how many permutations take at least 2 side by side S's.
Nosotros know from problem #ane there are 34,650 permutations of MISSISSIPPI and we now know that vii,350 arrangements take no adjacent Due south'due south, so to observe the permutations with at least 2 adjacent South'due south just accept the difference.
And then there are 27,300 permutations with 2 or more than side by side S's.
Thanks so much for reading!
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Source: https://medium.com/i-math/can-you-solve-the-mississippi-problem-6c0f3b02531
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