Answer to Question #140319 in Discrete Mathematics for Promise Omiponle

Question

How many strings of five uppercase English letters are there
(a) that start or end with the letters BO (in the order), if letters can be repeated? (inclusive or)
(b) that start with the letters BO (in that order), if letters can be repeated?
(c) that start and end with an X, if letters can be repeated?
(d) if letters can be repeated?

Answer 1

a) That start or end with the letters BO (in the order), if letters can be repeated

There are 26 possible letters.

First Latter = 26 way

Second Letter = 26 way

Third Letter = 26 ways

4th letter = 26 ways

5th letter = 26 ways

6th letter = 26 ways

7th letter = 1 ways (since it has to be B)

8th letter = 1 ways (since it has to be O)

Using the product rule :

$n_{end} with BO1*1*26^{6}=308,915,776$

b) That start with the letters BO (in that order), if letters can be repeated

There are 26 possible letters.

First Latter = 1 way (Since it has to be B)

Second Letter = 1 way (Since it has to be O)

Third Letter = 26 ways

4th letter = 26 ways

5th letter = 26 ways

6th letter = 26 ways

7th letter = 1 ways (Since it has to be B)

8th letter = 1 ways (Since it has to be B)

Using the product rule :

$1*1*26^{4}=456,976$

c) that start and end with an X, if letters can be repeated

There are 26 possible letters.

First Latter = 1 way (since it has to be X )

Second Letter = 26 way

Third Letter = 26 ways

4th letter = 26 ways

5th letter = 26 ways

6th letter = 26 ways

7th letter = 26 ways

8th letter = 1 ways (since it has to be X)

Using the product rule :

$1*1*26^{6}=308,915,776$

d) if letters can be repeated

There are 26 possible letters.

First Latter = 26 ways

Second Letter = 26 ways

Third Letter = 26 ways

4th letter = 26 ways

5th letter = 26 ways

6th letter = 26 ways

7th letter = 26 ways

8th letter = 26 ways

Using the product rule :

$26^{8}=208,827,064,576$