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# Possible three letter words

CCSS.Math:

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.

## Want to join the conversation?

- 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?(26 votes)
- 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.(1 vote)

- 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.(12 votes)- 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.(13 votes)

- 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! :)(8 votes)
- 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.(13 votes)

- How would you solve a problem that asked for all the three letter words with a vowel as the middle letter?(5 votes)
- 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`

(12 votes)

- At0:25, doesn't Sal mean "letters," and not "alphabets"?(10 votes)
- Yes, but in my opinion, the previous two corrections explain the third.(1 vote)

- How would u figure out actual three letter words rather than random letters, is there a way to eliminate those "non words"(4 votes)
- 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!(5 votes)

- 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?(3 votes)- 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.(5 votes)

- 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!(8 votes)

- What do I do if I want to find the permutations of all words that can be formed by the English alphabet, provided they are at least 3 letters long?

Edit: Maximum length of word is 26 letters(2 votes)- There are infinitely many words, since they can be as long as you like.(2 votes)

- The number of ways in which 9 persons can be divided into three equal groups is ?(2 votes)
- 1680. if you use the combinations formula: the first group of three people has 9C3 possibilities. for group #2, there are only 6 people left, so there are 6C3 combinations. the last group has 3C3 combinations. total number of ways to divide 9 people into 3 equal groups is 9C3*6C3*3C3=(9!/(3!6!))*(6!/(3!3!))*(3!/(3!0!))=84*20*1=1680. note:i didnt check my work because the numbers are kinda big.(2 votes)

## Video transcript

- [Voiceover] So let's ask ourselves some interesting questions about
alphabets in the English language. And in case you don't remember and are in the mood to count,
there are 26 alphabets. So if you go, "A, B, C,
D, E, F, G, H, I, J, K, L, M, N, O, P, Q, R, S,
T, U, V, W, X, Y, and Z," you'll get, you'll get 26, 26 alphabets. Now let's ask some interesting questions. So given that there are 26 alphabets in the English language, how many possible three letter words are there? And we're not going to be
thinking about phonetics or how hard it is to pronounce it. So, for example, the word, the word ZGT would be a legitimate
word in this example. Or the word, the word, the word SKJ would be a legitimate
word in this example. So how many possible three letter words are there in the English language? I encourage you to pause the video and try to think about it. Alright, I assume you've had a go at it. So let's just think about
it, for three letter words there's three spaces, so how many possibilities are there for the first one? Well, there's 26 possible
letters for the first one. Anything from a to z
would be completely fine. Now how many possibilities
for the second one? And I intentionally ask this to you to be a distractor because
we've seen a lot of examples. We're saying, "Oh,
there's 26 possibilities "for the first one and maybe
there's 25 for the second one, "and then 24 for the third,"
but that's not the case right over here because
we can repeat letters. I didn't say that all of the
letters had to be different. So, for example, the word, the word HHH would also be a legitimate word in our example right over here. So we have 26 possibilities
for the second letter and we have 26 possibilities
for the third letter. So we're going to have, and
I don't know what this is, 26 to the third power possibilities, or 26 times 26 time 26 and you
can figure out what that is. That is how many possible three letter words we can have for the English language if we didn't care about
how to pronounceable they are, if they meant anything
and if we repeated letters. Now let's ask a different question. What if we said, "How many
possible three letter words "are there if we want
all different letters?" So we want all different letters. So these all have to be different letters. Different, different
letters and once again, pause the video and see if
you can think it through. Alright, so this is where
permutations start to be useful. Although, I think a lot
of things like this, it's always best to reason through than try to figure out if
some formula applies to it. So in this situation,
well, if we went in order, we could have 26 different
letters for the first one, 26 different possibilities
for the first one. You know, I'm always
starting with that one, but there's nothing special
about the one on the left. We could say that the one on the right, there's 26 possibilities, well for each of those possibilities, for
each of those 26 possibilities, there might be 25 possibilities for what we put in the middle one if we say we're going to
figure out the middle one next. And then for each of these
25 times 26 possibilities for where we figured
out two of the letters, there's 24 possibilities
because we've already used two letters for the last
bucket that we haven't filled. And the only reason I went 26, 25, 24 is to show you there's nothing
special about always filling in the left most letter or
the left most chair first. It's just about, well,
let's just think in terms of let's just fill out
one of the buckets first. Hey, we have the most
possibilities for that. Once we use something up,
then for each of those possibilities we'll
have one left, one less for the next, the next bucket. And so I could do 24 times 25 times 26, but just so I don't fully confuse you, I'll go back to what I have been doing. 26 possibilities for the left most one. For each of those, you
would have 25 possibilities for the next one that you're
going to try to figure out because you already used one letter and they have to be different. And then for the last
bucket, you're going to have 24 possibilities, so this is going to be 26 times 25 times 24, whatever that happens to be. And if we wanted to
write it in the notation of permutations, we would
say that this is equal to, we're taking 26 things, sorry, not two p. 20, my brain is malfunctioning. 26, we're figuring out how
many permutations are there for putting 26 different things into three different spaces and this is 26, if we just blindly apply the formula, which I never suggest doing. It would be 26 factorial over 26 minus three factorial, which would be 26 factorial over 23 factorial, which is going to be
exactly this right over here because the 23 times 22 times
21 all the way down to one is going to cancel with the 23 factorial. And so the whole point of this video, there's two points, is one,
as soon as someone says, "How many different
letters could you form" or something like that, you don't just blindly do permutations or combinations. You think about well, what is
being asked in the question. Here, I really just have to
take 26 times 26 times 26. The other thing I want to point out, and I know I keep pointing it out, and it's probably getting tiring to you, is even when permutations are applicable, in my brain, at least,
it's always more valuable to just try to reason through the problem as opposed to just saying,
"Oh there's this formula "that I remember from weeks or years ago "in my life that had an N
factorial and K factorial "and I had to memorize
it, I have to look it up." Always much more useful
to just reason it through.