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Course: Precalculus>Unit 8

Lesson 3: Permutations

Possible three letter words

Sal explains how to find all of the possible three letter words when we can use each letter as many times as we want, and when each letter can only be used once.

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• This is not much relevant to the video, but what if the question is like this: how many ways are there to arrange L, L, L, B, B, D, D, D? How to use factorial to solve that?
• This something which is called permutation with repetition. You use this when you are interested in an arangement but you have elements which are similar.

You still need n elements (L, L, L, B, B, D, D, D) so n = 8.
Then you divide this by the factorial of similar elements. So k1 = 3 because you have 3 Ls. You need to do this for every differnt kind of element.
so k2 = 2 (2 Bs)
k3 = 3 (3 Ds)

Finally the formula is n!/k1!*k2!*k3!

Basically, you get the permutations of those 8 letters then rule out those options which are essentially the same. They aresame ina manner that if you swap the two Bs that seems like it is still the same.
• A permutation is interesting regardless of getting confused especially by varying questions being asked. Now I want to ask about how many distinct permutations can be made from the letters of the word MATH? and how many of these permutations starts with the letter M?

To solve it I just used 4!=24 distinct permutation if I'm correct then how many of these permutations start with M?
What I did is n!/n so 4!/4 = 6
I divided it by 4 since there are 4 distinct letters. Am I doing it right?
I need help since my math skills aren't that good.
• Yep. You got it.

Alternatively, the problem could also be rephrased as "how many permutations of A, T, and H are there?" (because M is fixed), so you'd still arrive at the answer of 3! = 6.
• What if you want all of the three-letter words that don't use the same letter twice, and you want the middle letter to be a vowel? You might start on the first space and write 26. Then the next space(the vowel) could be a 5 or a 6, depending on what the first letter was. If it was a vowel, it would make the middle space 5, and if it was a consonant, it would make the middle space 6. Also, which space you start on can change the final outcome. If you start on the middle space(the vowel), then it would always be 6, the next space would be 25, and the last space would be 24. May someone please explain this? Thanks! :)
• It might seem like you end up with different counts depending on where you start, but that is not actually the case. Lets breakdown what actually happens when you start with the first letter.

If you start with a consonant (and are counting y as a vowel) you have 20 choices for the first spot and then 6 choices for the middle slot, leaving 24 for the final slot.
If you instead start with a vowel, you have 6 choices for the first slot, then only 5 for the middle slot, and again 24 for the last.
So the total number of possibilities is 20*6*24 + 6*5*24 = (20*6 + 6*5)*24 = ((20+5)*6*24 =25*6*24 which is exactly what you had when you started with your vowel.
• How would you solve a problem that asked for all the three letter words with a vowel as the middle letter?
• Since the problem only mentions the need for a vowel to be in the middle, there's no restriction that vowel can't be in the first or the last positions.

So, for repeatable alphabets:
26*5*26

And for non-repeatable alphabets:
25*5*24
• At , doesn't Sal mean "letters," and not "alphabets"?
• Yes, but in my opinion, the previous two corrections explain the third.
• How would u figure out actual three letter words rather than random letters, is there a way to eliminate those "non words"
• That is a great question - but outside the scope of what we are doing here.
Checking syntax and formation of words is a whole other branch of study - and very language dependent.
Maybe the easiest would be to hand potential words over to a dictionary program for verification. - but if you are looking for a mathematical way to do it, you might be able to design a scoring function that biases certain letter combinations as (un)likely with simple rules, Let's let V = vowel and C = consonant - then perhaps the combination VCV (eg "age" or "ego") or CVC (eg "dad" or "mud") rate higher than CCV (are there any?). CCC and VVV combinations are ruled out, and some consideration given to CVV (eg "bee" or "moo") and VCC (eg "and" or "art" or "ebb" or "egg"). And what about VVC? ("aah" and "oil"!).
If you think about it enough, maybe it will be you who can propose a mathematical solution to the "three letter word problem."
Keep Studying!
• I tried doing my full name, lets say its Jenny Jiang, rewriting the full name in each scenario. I know it would be 10! on the top of the fraction, but I am confused on the bottom. Since my full name has 2 J's and 3 N's, it is confusing. Why would the bottom be 3!*2! ? why couldn't it be addition? Can someone prove this deductively?
(1 vote)
• Let's say for a moment that the J's and the N's were labelled, so we thought of them as different. That is, your name was spelled J1-E-N1-N2-Y-J2-I-A-N3-G, so that there were 10 "different" letters in your name. In that case, like you said, there would be 10! different permutations of the 10 letters, since, for example, J1-E-N1-N2-Y-J2-I-A-N3-G and J2-E-N1-N2-Y-J1-I-A-N3-G would be different (I switched the J1 and J2). Now, imagine that ENTIRE list of 10! (= 3628800) permutations (maybe you paid your little brother to write the entire list :) ), and imagine deciding, "Oh, actually, J1 and J2 should both just be J". You would find that in your list, suddenly every entry would be in the list TWICE, once from using J1 then J2, and another from using J2 then J1. For example, the pair above would BOTH become: J-E-N1-N2-Y-J-I-A-N3-G, and you would need to go through and erase exactly HALF of the permutations, so you would be at (10!)/2.

Now, with your list down to (10!)/2 (= 1814400) entries, imagine you decide the same about the N's: N1 and N2 and N3 all get changed to just N. When you look through the list now, you'll see that every entry will show up exactly 3! (= 6) times:
J-E-N1-N2-Y-J-I-A-N3-G --> J-E-N-N-Y-J-I-A-N-G
J-E-N1-N3-Y-J-I-A-N2-G --> J-E-N-N-Y-J-I-A-N-G
J-E-N2-N1-Y-J-I-A-N3-G --> J-E-N-N-Y-J-I-A-N-G
J-E-N2-N3-Y-J-I-A-N1-G --> J-E-N-N-Y-J-I-A-N-G
J-E-N3-N1-Y-J-I-A-N2-G --> J-E-N-N-Y-J-I-A-N-G
J-E-N3-N2-Y-J-I-A-N1-G --> J-E-N-N-Y-J-I-A-N-G

These will ALL become J-E-N-N-Y-J-I-A-N-G, and likewise every single entry will be repeated 6 times when N1, N2, and N3 are all changed to N, since there are 3! ways to put the three "different" N's into three spots. So now you will have to erase 5 out of every 6 entries, and the number of entries will become [(10!)/(2)]/6 = (10!)/(3!*2!). Hope that helps!
• This doesn't relate to the video much, but what if I have a question like...
How many arrangements are there using the letters in KITCHEN if the vowels must not be side by side? how would I solve that?
• I think that the solution is: 7!-2*6!=3600
7! is the number of possible arrangements without any limitation.
If we consider the vowels I & E side by side as if one letter, there is 6! possible arrangements multiplied by 2 for the two options IE or EI.