In this problem, a \"word\" is any string of seven (7) letters of the English al
ID: 3808475 • Letter: I
Question
In this problem, a "word" is any string of seven (7) letters of the English alphabet, with repeated letters allowed. Also, we will consider "Y" to be a vowel. Provide a brief justification for each of your answers (no more than five lines or so), explaining which counting rules you used and what your thought process was. Finally, feel free to have expressions of the form 3 middot 526, etc. in your final answers; no need to use calculators to compute such powers. 1. How many words are there? 2. How many words end with the letter T? 3. How many words begin with R and end with T? 4. How many words begin with A or B? 5. How many words begin with A or end with B? 6. How many words begin with A or B and end with A or B? 7. How many words begin with A or B or end with A or B? 8. How many words begin with AAB in some order? 9. How many words have no vowels? 10. How many words have exactly one vowel?Explanation / Answer
5. A word has 7 letters. So, first letter is fixed or the last letter is fixed while the other 6 letters can be in any combination. So, the number of words that begin with A or end with B are (26^6) + (26^6) i.e., there can be 26^6 combinations of words starting with A and 26^6 combinations of words ending with B.
6. Here the first letter can any of the two letters A and B. The last letter of the word also can be any of the two letters A and B. So, the number of words that begin with A or B and end with A or B is 2 * (26^5 )* 2 .
7. Here the condition is 'OR'. So, we need to add up the combinations seperately as follows:
Number of words that start with A or B is 2*(26^6)
Number of words that end with A or B is (26^6)*2
Add them to get the total number of words i.e., 2*(26^6) + (26^6)*2
8. The words should begin with AAB. Hence the first 3 letters of the word are fixed. So, the remaining 4 letters of the word can be any of the 26 letters. So, the total number of words that start with AAB are 26^4.
9. There are ftotally 6 vowels (including Y as said above). So, the number of letters without the vowels is 26-6=20. Now these letters can be of any order. So, the total number of the words that are not having vowels is 20^7.
10. The number of words having exactly one vowel is 6 * 20^6 because,
One letter can be a vowel. So, one letter can be any of the 6 vowels. The remaining letters are other than vowels. So, the number of letters remaining are 20. These 20 letters can be anywhere in the 6 positions of the word. So, number of words having exactly one vowel is 6 * 20^6.
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