The people living on the island of Gibberish use the standard Kobish alphabet ( letters, A through M). 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?
Hi Melody: How would you approach this problem? Here is my thinking: A,B,C,D,E,F,G,H,I,J,K,L,M (13 letters). I'm thinking of leaving out the letter "A" and permute the other 12 letters as: 12 P 3 =1,320, 3-letter permutations. Or, should I compute 13 P 3 =1,716 3-letter permutations? Then simply append the letter "A" at the beginning or the end of each permutation, which would give 4-letter "words" and yet would NOT change the number of permutations. That would leave the number of permutations unchanged, whether it is 1,320 or 1,716, except that each permutation would now be a 4-letter "word" !
What do you think of this scheme? I value your opinion. Thanks.
My reasoning is as follows:
There are 13^4 possible four letter words in total that can be made with 13 letters.
There are 12^4 possible four letter words that can be made with the 12 letters that aren't A.
Hence there are 13^4 - 12^4 four letter words that contain at least one A. (13^4 - 12^4 = 7825).
We then need to use similar reasoning for one two and three letter words.
Thanks Alan. Great idea! So:
13^4 - 12^4 =7,825 - 4-letter "words" that have at least one "A"
13^3 - 12^3 = 469 - 3-letter "words" that have at least one "A"
13^2 - 12^2 = 25 - 2-letter "words" that have at least one "A"
13^1 - 12^1 = 1 - 1-letter "word" that has one "A"
7,825 + 469 + 25 + 1 =8,320 "words" that have at least one "A"