Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Each runner during a race is labeled with a unique [#permalink]

Show Tags

16 Sep 2004, 20:54

00:00

A

B

C

D

E

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct
100% (00:02) wrong based on 2 sessions

HideShow timer Statistics

Each runner during a race is labeled with a unique one-letter code or a unique two-letter code, where the two letters are different. The codes use the 26-letter English alphabet. What is the maximum number of runners that can receive unique codes for the race?

676
26 + 26^2 - 26 = 26^2 = 676
First 26 is just the 26 individual letters
26^2 is for all possible combinations of 2 alphabet letters
Last 26 is unfavorable outcomes when 2 letters are similar ie AA,BB,CC... There are 26 of these possibilities _________________

Each runner during a race is labeled with a unique one-letter code or a unique two-letter code, where the two letters are different. The codes use the 26-letter English alphabet. What is the maximum number of runners that can receive unique codes for the race?

26 unique one-letter codes
(26*25) unique two letter codes where two letters are different = 650

total no. of alphabets = 26 pattern of code: 12 At place 1 we can have null or 26 characters. At place 2 we can have 26 characters or 25 characters(depending if null was at 1 or not) total becomes 1*26 + 26*25 = 26+ 650 = 676

or faster way is to start from right side at place 2, we have 26 characters at place 1 we have 25 character +1 null= 26 26*26=676

Unique letter combinations are 26 two letter combinations are = 26c2 = 26X25/2 = 13X25 = 325 two letter code where two letters are same = 26

total combinations = 26+325-26 = 325

answer should be 325

Would you say AC and CA are different codes or same? I think you will agree that they are different codes. Hence you do not use the combination formula here (26C2) because it only selects 2 different letters out of the 26 letters.

So how do you solve this question? You can do it in two different ways:

Method 1: You select 2 different letters and then arrange them. 26C2 * 2! = 650

Unique letter combinations are 26 two letter combinations are = 26c2 = 26X25/2 = 13X25 = 325 two letter code where two letters are same = 26

total combinations = 26+325-26 = 325

answer should be 325

Would you say AC and CA are different codes or same? I think you will agree that they are different codes. Hence you do not use the combination formula here (26C2) because it only selects 2 different letters out of the 26 letters.

So how do you solve this question? You can do it in two different ways:

Method 1: You select 2 different letters and then arrange them. 26C2 * 2! = 650

The number of one letter codes is 26 so total = 650 + 26 = 676

I agree with you and understood your point. however in one of kaplan CAT the OA is porvided as 351. Please see the attachment. Please anyone explain why OA is 351

Unique letter combinations are 26 two letter combinations are = 26c2 = 26X25/2 = 13X25 = 325 two letter code where two letters are same = 26

total combinations = 26+325-26 = 325

answer should be 325

Would you say AC and CA are different codes or same? I think you will agree that they are different codes. Hence you do not use the combination formula here (26C2) because it only selects 2 different letters out of the 26 letters.

So how do you solve this question? You can do it in two different ways:

Method 1: You select 2 different letters and then arrange them. 26C2 * 2! = 650

The number of one letter codes is 26 so total = 650 + 26 = 676

I agree with you and understood your point. however in one of kaplan CAT the OA is porvided as 351. Please see the attachment. Please anyone explain why OA is 351

Attachment:

Kaplan CAT.PNG

Yes, because the question in the attachment is different from the main post. It states an additional condition that if a 2 letter code is used, the reverse cannot be used.

So, AB- Is one valid code BA- will not be a valid code because AB has already been used.

Whereas according to the main question, both AB and BA will be considered valid codes.

Thanks fluke. This is the precise reason I opened a saparate topic as combinatorics-kaplan-700-level-ps-unique-2-latter-code-122067.html. This was locked by the 144144. can you please open the thread so that discussion can go forward _________________

Each runner during a race is labeled with a unique one-letter code or a unique two-letter code, where the two letters are different. The codes use the 26-letter English alphabet. What is the maximum number of runners that can receive unique codes for the race?

Ans- 676

Use slot method to solve this question:

For unique one letter code: 26 For unique two letter code: 26*25 = 650

Total # of codes = 26 + 650 = 676 _________________

GMAT is an addiction and I am darn addicted

Preparation for final battel: GMAT PREP-1 750 Q50 V41 - Oct 16 2011 GMAT PREP-2 710 Q50 V36 - Oct 22 2011 ==> Scored 50 in Quant second time in a row MGMAT---- -1 560 Q28 V39 - Oct 29 2011 ==> Left Quant half done and continued with Verbal. Happy to see Q39

MBA Admission Calculator Officially Launched After 2 years of effort and over 1,000 hours of work, I have finally launched my MBA Admission Calculator . The calculator uses the...

Final decisions are in: Berkeley: Denied with interview Tepper: Waitlisted with interview Rotman: Admitted with scholarship (withdrawn) Random French School: Admitted to MSc in Management with scholarship (...

The London Business School Admits Weekend officially kicked off on Saturday morning with registrations and breakfast. We received a carry bag, name tags, schedules and an MBA2018 tee at...

I may have spoken to over 50+ Said applicants over the course of my year, through various channels. I’ve been assigned as mentor to two incoming students. A...