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?

ANotSmartPerson
Oct 18, 2018

#1**+1 **

\(\text{There are }\dbinom{4}{1}19^3 \text{ different words containing a single instance of the letter A} \\ \dbinom{4}{1} \text{ spots for the A and then a choice of 19 letters, B-T, for each of the 3 remaining spots}\\ \text{Similarly there are }\dbinom{4}{k}19^{4-k} \text{ words containing }k \text{ instances of the letter A}\\ \text{Thus we have}\\ N=\sum \limits_{k=1}^4 ~\dbinom{4}{k}19^{4-k} = 29679\)

It strikes me that perhaps a simpler way of doing this is to subtract all the words with no instances of the letter A

from the total number of words possible.

\(\text{There are }19^4 \text{ words that have no A in them}\\ \text{There are }20^4 \text{ total possible words}\\ N=20^4-19^4 = 29679\)

Rom
Oct 18, 2018

#2**0 **

Thank you for trying but apparently the answer was wrong, here is the solution I got back:

We consider the opposite; we try to find the number of words that do not contain A, and then subtract it from the total possible number of words. So we have a few cases to consider:

One letter words: There is only 1 one-letter word that contains A, that's A.

Two letter words: There are 19*19=361 words that do not contain A. There is a total of 20*20=400 words, so we have 400-361=39 words that satisfy the condition.

Three letter words: There are 19*19*19=6859 words without A, and there are 20^3=800 words available. So there are 1141 words that satisfy the condition.

Four letter words: Using the same idea as above, we have 20^4-19^4 words satisfying the requirement.

So this gives a total of 1+39+1141+29679=**30860** words.

ANotSmartPerson
Oct 18, 2018