Let's break this up into cases. Ok. First, let's look at the number of words that are exactly 4 letters long. Right. So, one of the letters must be an A. There are 4 choices for that. After we pick where the A is going to go, we have to pick what the other 3 letters are going to be. For each of those 3 letters, we have 20 choices. Right. Now let's move on to words that are exactly 3 letters long. Ok. We already know that one of the letters has to be A. After we pick the A, we have to pick the other two letters For each of those two letters, we have 20 choices Let's move onto words that are exactly 2 letters long. Again, we already know that one of the letters is A. So we just have to pick one other letter. Right. There are 20 choices for that. So, there are $2(20)$ words that are exactly 2 letters long and contain the letter A. What about words that are exactly 1 letter long? Those are just the A's. There is only 1 one-letter long word that contains the letter A. Ok. So in total, there are $4(20^3) + 3(20^2) + 2(20) + 1$ words that contain the letter A. Let's simplify that. $4(20^3)$ is $4(8000)$, which is $32000$. Right. And $3(20^2)$ is $3(400)$, which is 1200. So the expression simplifies to $32000 + 1200 + 40 + 1$ Ok. Let's look at words that are 4 letters long. In order for a 4-letter word not to contain the letter A, none of the letters can be A. Right. So there are 19 choices for each letter. Right. So the number of 4-letter words that do not contain the letter A is $19^4$ Ok. There are $20^4$ total 4-letter words. So the number of 4-letter words that contain the letter A is $20^4 - 19^4$ Now let's move on to 3-letter words. In order to not contain the letter A, all 3 letters must be chosen from the 19 letters that are not A. So there are $19^3$ 3-letter words that do not contain the letter A And thus there are $20^3 - 19^3$ 3-letter words that contain the letter A. Right. Now let's move to 2-letter words. In order to not contain the letter A, both letters must be chosen from the 19 letters that are not A. So there are $19^2$ 2-letter words that do not contain the letter A. Therefore, there are $20^2 -19^2$ 2-letter words that contain the letter A. And finally, there is only 1 one-letter word that contains the letter A. So we can add up all those cases to get the total number of words that contain the letter A. Right. So $20^4 - 19^4 + 20^3 - 19^3 + 20^2 - 19^2 + 1$ is the total number of words that contain the letter A. Ok. Let's simplify that. $20^4$ is $160000$ and $19^4$ is $130321$. So $20^4 - 19^4$ is $160000 - 130321$ which is $29679$. Ok. Now let's look at the next term, $20^3 - 19^3$. $20^3$ is $8000$ and $19^3$ is $6859$. So $20^3 - 19^3$ is $8000 - 6859$ which is $1141$. Let's keep going. $20^2 - 19^2$ is $400 - 361$ which is $39$. So the expression simplifies to $29679 + 1141 + 39 + 1$ And if we add all those terms we get $30860$ So there are $30860$ words that contain the letter A Ok. That's our final answer.


The inhabitants of the island of Jumble use the standard Kobish alphabet ($20$ letters, A through T).  Each word in their language is $4$ letters or less, and for some reason, they insist that all words contain the letter A at least once. How many words are possible?