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Once again, thanks Manish!
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Once again, thanks Manish!

You are welcome!! The short cuts of approximation and using the units digits are nice. Time saving.
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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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Three different characters: C1 C2 C3
Theory: When calculating the number of items it is (xn-xa)+1

C1
8-1+1=8

C2
25th letter is y; 2nd letter is b
25-2+1=24

C3
8+24=32

Theory: Fundamental counting rule C1·C2·C3
8·24·32

I don't like multiplying 3 numbers together. I split the 8 up
(4·24)·(2·32)=96·64

Theory: the units digits of the answer results only from the units digits of the inputs
6·4=x4, so we know the answer ends in 4

B is too small. D must be the answer
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Hello from the GMAT Club BumpBot!

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