SOLUTION: How many strings of five upper class English letters are there for each of the following restrictions? A string of letters need not be a real word, so XYZWT is a good string. a

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Question 459567: How many strings of five upper class English letters are there for each of the following restrictions?
A string of letters need not be a real word, so XYZWT is a good string.
a. Start with an X and letters can be repeated
b. No letter can be repeated
c. Start and end with an X and letters can be repeated
d. Start with the letters BO(in that order) and letters can be repeated
e. Start and end with the letters BO(in that order) and letters can be repeated
f. Start or end with the letters BO(in that order)if letters can be repeated
Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
a. Start with an X and letters can be repeated 
Examples:  XAJQN, XLLTX, XXXXX 

Choose letter#1 1 way, choose letter#2 26 ways, 
choose letter#3 26 ways, choose letter#4 26 ways,
choose letter#5 26 ways.

That's 1×26×26×26×26 = 456976 ways

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b. No letter can be repeated
Examples:  kYUDM, PIANO, VWXYZ

Choose letter#1 26 ways, choose letter#2 25 ways, 
choose letter#3 24 ways, choose letter#4 23 ways,
choose letter#5 22 ways.

That's 26×25×24×23×22 = 7893600 ways.

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c. Start and end with an X and letters can be repeated 
Examples:  XEROX, XFFJX, XXXXX

Choose letter#1 1 way, choose letter#2 26 ways, 
choose letter#3 26 ways, choose letter#4 26 ways,
choose letter#5 1 way.

That's 1×26×26×26×1 = 17576 ways.

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d. Start with the letters BO(in that order) and letters can be repeated 
Examples:  BOOZE, BOAST, BOBOO

Choose letter#1 1 way, choose letter#2 1 way, 
choose letter#3 26 ways, choose letter#4 26 ways,
choose letter#5 26 ways.

That's 1×1×26×26×26 = 17576 ways.

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e. Start and end with the letters BO(in that order) and letters can be repeated 
Examples:  BOOBO, BOMBO, BOBBO

Choose letter#1 1 way, choose letter#2 1 way, 
choose letter#3 26 ways, choose letter#4 1 way,
choose letter#5 1 ways.

That's 1×1×26×1×1 = 26 ways.
 
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f. Start or end with the letters BO(in that order)if letters can be repeated
We must use a formula here:

N(X or Y) = N(X) + N(Y) - N(X and Y), 
  where N( ) means "the number of".

Let X be "words beginning with BO", 
        calculated in part d, so N(X) = 17576
Let Y be "words ending with BO", 
        same as the results of part d, so N(Y) = 17576
Then "X and Y" will be "words beginning 
        and ending with BO", calculated 
        in part e, so N(X and Y) = 26

N(X or Y) = 17576 + 17576 - 26 = 35126

Edwin