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100 points for $49 worth of Veritas practice GMATs FREE VERITAS PRACTICE GMAT EXAMS Earn 10 Points Per Post Earn 10 Points Per Thanks Earn 10 Points Per Upvote ## A 3-character alpha-numeric code does have the following ##### This topic has 2 expert replies and 0 member replies ### Top Member ## A 3-character alpha-numeric code does have the following ## Timer 00:00 ## Your Answer A B C D E ## Global Stats Difficult 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 ### GMAT/MBA Expert GMAT Instructor Joined 25 Apr 2015 Posted: 2852 messages Followed by: 18 members Upvotes: 43 Top Reply 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 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: 8 x 24 x 32 = 6,144 Answer: D _________________ Scott Woodbury-Stewart Founder and CEO scott@targettestprep.com See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews ### GMAT/MBA Expert GMAT Instructor Joined 25 May 2010 Posted: 15362 messages Followed by: 1866 members Upvotes: 13060 GMAT Score: 790 Top Reply 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 first character = 8. (Of the 10 digits, any digit but 0 or 9.) 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. _________________ Mitch Hunt Private Tutor for the GMAT and GRE GMATGuruNY@gmail.com If you find one of my posts helpful, please take a moment to click on the "UPVOTE" icon. Available for tutoring in NYC and long-distance. For more information, please email me at GMATGuruNY@gmail.com. Student Review #1 Student Review #2 Student Review #3 Free GMAT Practice Test How can you improve your test score if you don't know your baseline score? Take a free online practice exam. Get started on achieving your dream score today! Sign up now. • Free Trial & Practice Exam BEAT THE GMAT EXCLUSIVE Available with Beat the GMAT members only code • 5-Day Free Trial 5-day free, full-access trial TTP Quant Available with Beat the GMAT members only code • Free Veritas GMAT Class Experience Lesson 1 Live Free Available with Beat the GMAT members only code • Award-winning private GMAT tutoring Register now and save up to$200

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