BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Target Test Prep · GMAT

Choose how you want to prepare

Learn live with an expert or move at your own pace. Every option includes the complete TTP study system.

★★★★★5.0559 reviews
GMATLiveTeach 7 seats left
Chris Peckover
NEXT LIVE COHORT

Oct 13 to Jan 7, 2027

with Chris Peckover

Schedule
Tue, Thu · 8:00 to 10:00 PM ET
Included
40 live hours + 6 months of GMAT OnDemand
  • Live instruction and real-time questions
  • Class recordings and assigned practice
View class & enroll
Limited cohort · enrollment openTarget Test Prep
EALiveTeach 5 seats left
Logan Thompson
EXECUTIVE ASSESSMENT

Sep 6 to Dec 6, 2026

with Logan Thompson

Schedule
Sun · 9:30 AM to 12:30 PM ET
Included
Live EA class + 6 months of EA OnDemand
  • Expert-led weekly online sessions
  • EA Masterclass access between classes
View EA class & enroll
Limited cohort · enrollment openTarget Test Prep
GMATOnDemand Start anytime
SELF-PACED MASTERCLASS

Target Test Prep GMAT OnDemand

Complete access from day one. Study on your schedule.

130-point score guarantee
$0to start then $127/mo
  • Personalized study plan and analytics
  • Thousands of lessons and practice questions

Compare the format, schedule, and included access before enrolling. Prices and seat counts shown reflect the supplied offer details.

probability and alphabets

Expert replies
Source: — Problem Solving |

by Neo2000 » Fri Jan 09, 2009 11:17 am
If you are creating 1 Letter Code you can get 26 codes
Two letter codes with repetition allowed = 26 x 26 = 576
Three Letter codes = 26 x 26 x 26 = 26^3

You know that any power of 6 ends in a 6
And the Units digit of all 3 numbers is a 6
Which means your answer must end in an 8
Join the discussion

Choice problem

by katz » Fri Jan 09, 2009 11:44 am
Let us start by spelling out the rule of counting:
The rule of counting says that if we have m ways of selecting first item, n ways of selecting second item, o ways of selecting third item and so on then we have a total of m*n*o*... total ways of selecting items.

The second thing that we should be aware of in (probability) problems of these types is whether the problem calls for "or" or "and" operation. In probability theory when you see the word "or" it is an indication for you to use "+" and when you see "and" you apply "*".

This problem calls for "or" indicating us to sum up the numbers.

So far so good. Now let us get into the math.

1. Three letter coded stock: Imagine three placeholders _, _, _. Now the first place can be filled with any alphabets from A to Z in 26 ways. Also since we are free to repeat the alphabets, we would have another 26 ways of filling the second placeholder, and continuing in this fashion, we can fill the third place in another 26 ways. So the total numbers of choices we have for three letter coded stock is 26*26*26 = 17,576

2. Two letter coded stock: Here we have two placeholders _,_. Extending the logic from point 1 above, we would have 26*26 = 676 choices.

3. One letter coded stock: Here we have one placeholder _, and so only 26 choices.

Final Answer: Now since the problem read "...one-, two-, or, three...", we would need to sum (+) up the numbers i.e. 17,576+676+26 = 18,278 choices.

So "E" is the correct answer choice.
Join the discussion