Asked by ashley
How many possible license plates are possible in which two letters are followed by at least 1 digit and no more than 4 digits?
Answers
Answered by
MathMate
For the two letters, there are 26²=676 arrangements.
For the digits, it is the sum of 4 cases, namely:
1 digit = 10 arrangements
2 digits = 100 arrangements
3 digits = 1000 arrangements
4 digits = 10000 arrangements
for a total of 11110 arrangements.
Since the letters and digits are independent of each other, the total number of arrangements is the product of the two, namely:
676*11110 arrangements.
Note:
For a 0 to n-character string with an alphabet of k, the number of arrangements is (k<sup>n+1</sup>-1)/(k-1).
For the case in question, we first calculate 0 to 4 character string with an alphabet of 10 digits,
N=(10<sup>4+1</sup>-1)/(10-1)
=(100000-1)/(10-1)
=11111
But since 0 digit is excluded, we subtract 1 to get 11110.
For the digits, it is the sum of 4 cases, namely:
1 digit = 10 arrangements
2 digits = 100 arrangements
3 digits = 1000 arrangements
4 digits = 10000 arrangements
for a total of 11110 arrangements.
Since the letters and digits are independent of each other, the total number of arrangements is the product of the two, namely:
676*11110 arrangements.
Note:
For a 0 to n-character string with an alphabet of k, the number of arrangements is (k<sup>n+1</sup>-1)/(k-1).
For the case in question, we first calculate 0 to 4 character string with an alphabet of 10 digits,
N=(10<sup>4+1</sup>-1)/(10-1)
=(100000-1)/(10-1)
=11111
But since 0 digit is excluded, we subtract 1 to get 11110.
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