Question 1042201: Please help me on parts A through C its confusing.
4 A DE license plate starts with a digit other than 0, followed by two capital letters, followed by three more digits (0 through 9).
a) How many different DE plates are possible? ____________________________
b) How many different DE plates start with a 4 and end with a 9? _____________________________
c) How many DE plates have no repeated symbols? __________________________________
(all digits and letters are different)
Answer by Aldorozos(172)   (Show Source):
You can put this solution on YOUR website!
The Plan number will look like something like this
NAANNN
The first N above is a number between 1 and 9. For this N we have nine options. These options are 1,2,3,4,5,6,7,8,9
For the first A above we have 26 options. These options are A,B,C,....Z
For the Second A above we have 26 options. These options are A,B,C,....Z
For the second N which is a number we can have 10 options. These options are
0,1,2,3,4,5,6,7,8,9
For the third N which is a number we can have 10 options. These options are
0,1,2,3,4,5,6,7,8,9
for the fourth N which is a number we can have 10 options. These options are
0,1,2,3,4,5,6,7,8,9
Now the question a asks how many plate numbers we can have. We can multiply our options
9*26*26*10*10*10 = the answer to question a. You may use your calculator for fine this number
Questions b is asking
How many different DE plates start with a 4 and end with a 9?
This means for the first number we don't have nine different options anymore. We only have one option and that option is number 4.
Similarly for the last digit we don't have 10 different options. We only have one option and that option is 9
1*26*26*10*10*1 = the answer to question b. You may use your calculator for fine this number
Question C doesn't want any repeated numbers.
For the first digit we have nine options 1,2,3,4,5,6,7,8,9
For the first alphabet we have 26 options
However we come with the second alphabet, we don't have 26 options anymore we have 25 because we don't want repeated alphabet for the second alphabet.
For the second digit that we originally had 10 options, this time we have 9 options since we already used one of the options in the first digits
For the third option we only have 8 digits left. Since we used one of the digits in the first option and another digit (number) in the second option. Therefore we have ended up with only 8 numbers
For the third last number we have 7 options left.
Now we have to multiply out options together
9*26*25*9*8*7 This is the answer to C. You may use your calculator to find the answer.
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