The 3 character alpha-numeric code has the following characters

First character - Any number except 0 and 9 - Total of 10 - 2 = 8 possibilities
Second character - Any small letter except a and z - Total of 26 - 2 = 24 possibilities
Third character - Any number except 0 and 9 or Any small letter except a and z - Total of 8 + 24 = 32 possibilities

In this case, to find the total number of possibilities, we can use the unit's digit to find the
answer as number ending 8 * number ending 4 * number ending 2 is a number ending 4.
Also, the answer must be close to 10*20*30 is a little more than 6000.
P.S when the answer options are far apart, we can use approximations to arrive at the answer

Therefore, Option D(6144) is the the total number of 3-character alpha-numberic codes.
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Once again, thanks Manish!
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Solution

Given:

• The code is of 3-character alpha-numeric
• 1st character can be any number except 0 and 9
• 2nd character can be any small letter between a to z, excluded
• 3rd character can be any of those characters possible for the first two places

To find:

• Keeping all the constraints satisfied, how many such codes can be formed

Approach and Working:

• As the 1st character can be any number except 0 and 9, number of possibilities exist for the first character = 8
• As the 2nd character can be any small letter between a to z, excluded, number of possibilities exist for the second character = 24
• As the 3rd character can be any of those characters possible for the first two places, number of possibilities exist for the third character = 8 + 24 = 32
• Therefore, the total number of possible codes = 8 * 24 * 32 = 6144

Hence, the correct answer is option D.

Answer: D
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