Events & Promotions
|
|
GMAT Club Daily Prep
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
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Jun 2010:00 AM PDT
-11:00 AM PDT
Class is officially in session! Join Charles Bibilos from GMAT Ninja and GMAC’s own Jaime Malatesta, Senior Psychometrician, as they simplify some of the complexities of GMAT scoring.
Jun 2106:30 AM PDT
-08:30 AM PDT
Not only the typical inference questions but also all other CR questions on the GMAT require the ability to deduce from the given information. Come master how to draw inferences accurately and efficiently with Verbal Expert-Sunita Singhvi. Limited Seats!
Jun 2111:00 AM IST
-01:00 PM IST
Do RC/MSR passages scare you? e-GMAT is conducting a masterclass to help you learn – Learn effective reading strategies Tackle difficult RC & MSR with confidence Excel in timed test environment
Jun 2211:30 AM EDT
-12:30 PM EDT
Learn how Keshav, a Chartered Accountant, scored an impressive 705 on GMAT in just 30 days with GMATWhiz's expert guidance. In this video, he shares preparation tips and strategies that worked for him, including the mock, time management, and more.
Jun 2308:00 PM PDT
-09:00 PM PDT
Full-length FE mock with insightful analytics, weakness diagnosis, and video explanations!- Jun 30
08:00 AM PDT
-11:59 PM PDT
Join us June 30th for 2 weeks of GMAT questions (just 8 per day) to help with your prep and to compete with your fellow GMAT Clubbers for a chance to win a prize fund of $30,000 for you and your team-mates!
E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
51% (02:36) correct
49%
(02:35)
wrong
based on 1649
sessions
History
Date
Time
Result
Not Attempted Yet
A 5-digit code consists of one number digit chosen from 1, 2, 3 and four letters chosen from A, B, C, D, E. If the first and last digit must be a letter digit and each digit can appear more than once in a code, how many different codes are possible?
A. 375
B. 625
C. 1,875
D. 3,750
E. 5,625
A. 375
B. 625
C. 1,875
D. 3,750
E. 5,625
09 May 2012, 01:23
Kudos
Bookmarks
samarthgupta
A friend asked me this question recently and I wasn't able to get the official answer for this question. I am not sure about the source or the difficulty level. The questions is as follows :-
A 5-digit code consists of one number digit chosen from 1, 2, 3 and four letters chosen from A, B, C, D, E. If the first and last digit must be a letter digit and each digit can appear more than once in a code, how many different codes are possible?
A. 375
B. 625
C. 1,875
D. 3,750
E. 5,625
A 5-digit code consists of one number digit chosen from 1, 2, 3 and four letters chosen from A, B, C, D, E. If the first and last digit must be a letter digit and each digit can appear more than once in a code, how many different codes are possible?
A. 375
B. 625
C. 1,875
D. 3,750
E. 5,625
Notice that each digit can appear more than once in a code.
Since there should be 4 letters in a code (X-X-X-X) and each letter can take 5 values (A, B, C, D, E) then total # of combinations of the letters only is 5*5*5*5=5^4.
Now, we are told that the first and last digit must be a letter digit, so number digit can take any of the three slots between the letters: X-X-X-X, so 3 positions and the digit itself can take 3 values (1, 2, 3).
So, total # of codes is 5^4*3*3=5,625.
Answer: E.
Similar question to practice: a-4-letter-code-word-consists-of-letters-a-b-and-c-if-the-59065.html
Hope it helps.
09 May 2012, 09:55
Kudos
Bookmarks
samarthgupta
GyanOne
Choose one number from 3 numbers in 3C1 = 3 ways
Choose four letters from 5 letters in 5C4 = 5 ways
Choose one position from the middle three for the number in 3C1 = 3 ways
The other four positions can be filled by the 4 letters in 4^4 ways.
Therefore total number of codes possible = 3*5*3*(4^4) = 45*16*16 = 11,520
Choose four letters from 5 letters in 5C4 = 5 ways
Choose one position from the middle three for the number in 3C1 = 3 ways
The other four positions can be filled by the 4 letters in 4^4 ways.
Therefore total number of codes possible = 3*5*3*(4^4) = 45*16*16 = 11,520
For the first and last positions the letters can be chosen in 5c1 ways, but if we get say the same letter twice then we need to divide by 2! to avoid duplicates since we are looking for different codes.
For the 3 positions in between, we can choose 1 digit in 4c1 ways, and the remaining 2 digits in 5c1 and 5c1 ways. These three can rearrange themselves in 3! ways and we again divide by 2! to avoid duplicates since we can have the letters to repeat themselves. I have shown this below :-
5c1 [(4c1*5c1*5c1)3!]/2! 5c1
1 2 3 4 5
Therefore total according to me would be :-
[(5c1*5c1)/2!]*[(4c1*5c1*5c1)3!]/2! = (25*25*6*4)/4 = 3750.
Can anyone correct where i am making a mistake.
You are allowed duplicates. Even if A appears in the first as well as the last position, it will give you a code different from what you get when you have different letters in the first and the last position. You need to arrange the letters here. If instead you needed to just select groups, then yes, you would have worried about the effect of duplicates.
You select a letter for the first position in 5C1 ways and a letter for the last position in 5C1 ways.
Say you put the digit in the second position. You can select a digit for the second position in 3C1 ways.
You can select the letters for the third and fourth positions in 5C1 and 5C1 ways.
Hence, you get 5*3*5*5*5 codes.
But here, we have put the digit in the second place. It could have been in the third or fourth place too. So you multiply the above given result by 3.
Hence total number of codes = 5*3*5*5*5*3 = 5625
