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?
consisting of 1 letter ---- 1 , namely the A
consisting of 2 letters ::
without restrictions= 20x20 = 400
without the A = 19x19 = 361
at least 1 A ----> 400-361 = 39
consisting of 3 letters, no restricion = 20^3 = 8000
without any A's = 19^3 = 6859
at least one A = 8000-6859 = ....
do the 4 letter case the same way, add up the 4 cases.
To find out how many words are possible, we need to consider all the possible combinations of letters that satisfy the given conditions.
We know that each word must have at least one instance of the letter A. Since the words can have a maximum of 4 letters, let's consider the different scenarios:
1. Words with 1 letter: There is only one possibility, which is the letter A itself.
2. Words with 2 letters: In addition to the word "A", we can have 19 other letters to choose from to make the second letter. Therefore, there are 19 possible combinations in this scenario.
3. Words with 3 letters: Again, we can choose any one of the 19 letters for the first position, any of the 20 letters for the second position (including A), and any of the 20 letters for the third position. So, the total number of combinations in this scenario is 19 * 20 * 20.
4. Words with 4 letters: Similarly, we can choose any one of the 19 letters for the first position, any of the 20 letters for the second, third, and fourth positions. Therefore, the total number of combinations in this scenario is 19 * 20 * 20 * 20.
To find the total number of words possible, we sum up the possibilities from all the different scenarios:
Total = 1 + 19 + (19 * 20 * 20) + (19 * 20 * 20 * 20)
Calculating this expression, we find that the total number of words possible in the given scenario is 77,221 words.