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Martina1120

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Hi I am working on a problem as follows:

Students log on to the California Virtual Campus with a user name consisting of eight characters: four uppercase letters of the alphabet followed by four digits.

a. How many user names are theoretically possible for this system?
26*26*26*26*10*10*10*10= 4,569,760,000

b. How many user names have no repeated letters or digits?
26*25*24*23*10*9*8*7=1,808,352,000

this is where i got stuck:

c. How many user names have no matching adjacent letters or digits?I dont see the difference between this and question b? the solution is : 26*25*25*25*10*9*9*9=2,961,562,500 but why is just the second option repeated?

d. How many user names begin with A and end with 9? I don't even know where to start with this one.

Any help that anyone can provide would be greatly appreciated. I know this is simple stuff, but I'm just not catching on right away. Thanks!
 
Martina1120 said:
Hi I am working on a problem as follows:

Students log on to the California Virtual Campus with a user name consisting of eight characters: four uppercase letters of the alphabet followed by four digits.

a. How many user names are theoretically possible for this system?
26*26*26*26*10*10*10*10= 4,569,760,000

b. How many user names have no repeated letters or digits?
26*25*24*23*10*9*8*7=1,808,352,000

this is where i got stuck:

c. How many user names have no matching adjacent letters or digits?I dont see the difference between this and question b? the solution is : 26*25*25*25*10*9*9*9=2,961,562,500 but why is just the second option repeated?

Say you choose A for first letter (26 choices) - then the next letter cannot be A (25 choices - say you choose B) - now for the third letter A is free to be chosen (that was not the case in (b)) and you have 25 choices......
d. How many user names begin with A and end with 9? I don't even know where to start with this one.

For the first choice you have only one choice - 'A' and for the last space you have only one choice '9'. Rest is like (a)
Any help that anyone can provide would be greatly appreciated. I know this is simple stuff, but I'm just not catching on right away. Thanks!
Click to expand...
 
Hello, "Martina1120!

Students have a user name consisting: four uppercase letters of the alphabet followed by four digits.

c. How many user names have no matching adjacent letters or digits?
Click to expand...

The 1st character can be any letter: 26 choices.
The 2nd must not match the 1st one: 25 choices.
The 3rd must not match the 2nd one: 25 choices.
The 4th must not match the 3rd one: 25 choices.

The 5th can be any digit: 10 choices.
The 6th must not match the 5th: 9 choices.
The 7th must not match the 6th: 9 choices.
The 8th must not match the 7th: 9 choices.

Therefore, there are: .\(\displaystyle 26\cdot25\cdot25\cdot25\cdot10\cdot 9\cdot9\cdot 9\) ways.


d. How many user names begin with A and end with 9?
Click to expand...

The first character is \(\displaystyle A\) and the last is \(\displaystyle 9.\)

The user name looks like this: .\(\displaystyle A\:\_\;\_\;\_\;\_\;\_\;\_\;9\)
. . and we want three letters followed by three digits in the spaces.

There are: .\(\displaystyle 26\cdot26\cdot26\cdot10\cdot10\cdot10\) ways.

 
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