A 3-character alpha-numeric code does have the following properties - the first character can be any number except 0 and 9, the second character can be any small letter between a to z, excluded, and the third character can have any of those characters possible for the first two places. How many such codes can be formed?
A. 26
B. 64
C. 520
D. 6144
E. 9360
OA D
Source: e-GMAT
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A 3-character alpha-numeric code does have the following
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BTGmoderatorDC
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Number of options for the first character = 8. (Of the 10 digits, any digit but 0 or 9.)BTGmoderatorDC wrote:A 3-character alpha-numeric code does have the following properties - the first character can be any digit except 0 and 9, the second character can be any small letter between a and z, excluding a and z, and the third character can be any of those characters possible for the first two places. How many such codes can be formed?
A. 26
B. 64
C. 520
D. 6144
E. 9360
Number of options for the second character = 24. (Of the 26 letters in the alphabet, any but a or z.)
Number of options for the third character = (number of options for the first character) + (number of options for the second character) = 8+24 = 32.
To combine these options, we multiply:
8*24*32 = large integer with a units digit of 4.
The correct answer is D.
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Scott@TargetTestPrep
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There are 8 choices for the first digit, 26 - 2 = 24 choices for the second digit, and 24 + 8 = 32 choices for the last digit. The number of ways to create the code is:BTGmoderatorDC wrote:A 3-character alpha-numeric code does have the following properties - the first character can be any number except 0 and 9, the second character can be any small letter between a to z, excluded, and the third character can have any of those characters possible for the first two places. How many such codes can be formed?
A. 26
B. 64
C. 520
D. 6144
E. 9360
OA D
Source: e-GMAT
8 x 24 x 32 = 6,144
Answer: D
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