# Permutations / Combinations - suppose a word is a string of 8 letters of the alphabet with repeated letters allowed

1.) How many words are there?

Not sure how to solve this since repeated letters are allowed. $n^r$ is the formula we are told to use for permutations with repeated objects, but $26^8$ seems like too large of a number.

2.) How many words end with the letter T?

I assume this would be a combination of the form $26 \choose 1$

3.) How many words begin with R and end with T?

I assume this would be a combination of the form $26 \choose 2$

4.) How many words start with A or B?

I assume this would be a combination of the form $26 \choose 1$ $+$ $26 \choose 1$

5.) How many words begin with A or end with B?

Same as number 4?

## 2 Answers

1: $26^8$ is correct. There are 26 choices for the first letter, 26 for the second letter, etc. You multiply all the 26's because these choices are independent. "multiplication principle"

2: The last letter is known. Otherwise, you choose the first 7 letters freely, so it's $26^7$.

3: I leave for you.

4: This is the "addition principle". Count words beginning with A, and words beginning with B. These form two disjoint sets, as a word cannot begin with both A and B. Hence you add the two answers.

5: This is not the same as 4, because now it is possible to both begin with A and end with B. Such words were counted twice. Hence, you need to add the two answers, but now subtract the double-counted words. How many double-counted words are there? Such a word begins with A and ends with B, so there are 6 letters in the middle freely chosen.

• So for 4 would it be $26 \choose 7$ $+$ $26 \choose 7$? – hax0r_n_code Nov 17 '14 at 3:42
• And for 5 it would be $26 \choose 7$$26 \choose 7$? – hax0r_n_code Nov 17 '14 at 3:43
• None of these problems are solved with binomial coefficients. – vadim123 Nov 17 '14 at 4:06
• $26^7+26^7$ for number 4 and $26^7(26^7)$ for number 5 – hax0r_n_code Nov 17 '14 at 4:10
1. $26^8$ is correct. There are $26$ options for each of the $8$ letters.

2. $26^7$, since you can choose the first $7$ letters, but the last one has to be T.

3. $26^6$, since you can only choose the middle $6$.

4. You can choose the last $7$ letters. There are $2$ options for the first one. What do you get?

5. Do words beginning with A, ending with B, and both beginning with A and ending with B separately.

• Ahhh for 2 and 3 I was looking at it the wrong way. Thank you! – hax0r_n_code Nov 17 '14 at 3:23