+283 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?
+6248 \(\text{There are }\dbinom{4}{1}19^3 \text{ different words containing a single instance of the letter A} \\ \dbinom{4}{1} \text{ spots for the A and then a choice of 19 letters, B-T, for each of the 3 remaining spots}\\ \text{Similarly there are }\dbinom{4}{k}19^{4-k} \text{ words containing }k \text{ instances of the letter A}\\ \text{Thus we have}\\ N=\sum \limits_{k=1}^4 ~\dbinom{4}{k}19^{4-k} = 29679\)
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.
\(\text{There are }19^4 \text{ words that have no A in them}\\ \text{There are }20^4 \text{ total possible words}\\ N=20^4-19^4 = 29679\)
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+283 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.
