2
$\begingroup$

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?

$\endgroup$
5
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ So for 4 would it be $26 \choose 7$ $+$ $26 \choose 7$? $\endgroup$ – hax0r_n_code Nov 17 '14 at 3:42
  • $\begingroup$ And for 5 it would be $26 \choose 7$$26 \choose 7$? $\endgroup$ – hax0r_n_code Nov 17 '14 at 3:43
  • $\begingroup$ None of these problems are solved with binomial coefficients. $\endgroup$ – vadim123 Nov 17 '14 at 4:06
  • $\begingroup$ $26^7+26^7$ for number 4 and $26^7(26^7)$ for number 5 $\endgroup$ – hax0r_n_code Nov 17 '14 at 4:10
2
$\begingroup$
  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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.