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.
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:
Want to solve 700+ level Algebra questions within 2 minutes? Attend this free webinar to learn how to master the most challenging Inequalities and Absolute Values questions in GMAT
Each admissions season, many candidates receive a response from MBA admissions committees that can sometimes be far more frustrating than a rejection: “You have been placed on our waitlist.” What should you do when your status is uncertain?
Are you struggling to achieve your target GMAT score? Most students struggle to cross GMAT 700 because they lack a strategic plan of action. Attend this Free Strategy Webinar, which will empower you to create a well-defined study plan to score 760+
A 5-digit code consists of one number digit chosen from 1, 2
[#permalink]
Show Tags
08 May 2012, 23:30
4
65
00:00
A
B
C
D
E
Difficulty:
95% (hard)
Question Stats:
50% (02:40) correct 50% (02:31) wrong based on 822 sessions
HideShow timer Statistics
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?
Re: A 5-digit code consists of one number digit chosen from 1, 2
[#permalink]
Show Tags
09 May 2012, 01:23
10
6
samarthgupta wrote:
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
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).
Choose one number from 3 numbers in 3C1 = 3 ways Choose one position from the middle three for the number in 3C1 = 3 ways The other four positions can be filled by the 5 letters in 5^4 ways.
Therefore total number of codes possible = 3*3*(5^4) = 5,625
E
_________________
GyanOne [www.gyanone.com]| Premium MBA and MiM Admissions Consulting
Awesome Work | Honest Advise | Outstanding Results
Reach Out, Lets chat! Email: info at gyanone dot com | +91 98998 31738 | Skype: gyanone.services
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
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 :-
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
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 :-
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
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
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
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 :-
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
Karishma can you tell me wher I am wrong- 1st and last places can be filled in 5 X 5 ways, another 3 this way
5C1X5C1X3C1 and they can be arranged among 3! ways = 5C1X5C1X3C1 x3! = 5x5x3x3x2
Karishma can you tell me wher I am wrong- 1st and last places can be filled in 5 X 5 ways, another 3 this way
5C1X5C1X3C1 and they can be arranged among 3! ways = 5C1X5C1X3C1 x3! = 5x5x3x3x2
finally 5x5x5x5x3x3x2
I am not sure how you have worked this out.
5*5 is fine for first and last positions - they must be letters. But you need 5 digit code so you have another 3 positions to fill
* _ _ _ *
You must use one number digit so you can select a number digit in 3 ways and the position for that number digit in 3 ways. Also the other two positions must be letters so they can be selected in 5*5 ways.
In all, 5*5*3*3*5*5
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
Re: A 5-digit code consists of one number digit chosen from 1, 2
[#permalink]
Show Tags
12 May 2015, 05:36
Can someone pls tell me if my approach is correct :
We need to fill these 5 blank spaces. _ _ _ _ _ We have 5 letters of which we need to pick 4. That makes it 5P4. Now 4 of these 5 letters could repeat too. So to account for this, we need to divide 5P4 by 4! Of the 3 numbers we need to pick 1. So 3P1. This makes it 5P4/4! * 3P1 * 5P4/4! * 5P4/4! * 5P4/4! or 5P4/4! * 5P4/4! * 3P1 * 5P4/4! * 5P4/4! or 5P4/4! * 5P4/4! * 5P4/4! * 3P1 * 5P4/4!
Accounting for all the scenarios we get 5P4/4! * 3P1 * 5P4/4! * 5P4/4! * 5P4/4! + 5P4/4! * 5P4/4! * 3P1 * 5P4/4! * 5P4/4! + 5P4/4! * 5P4/4! * 5P4/4! * 3P1 * 5P4/4!
Re: A 5-digit code consists of one number digit chosen from 1, 2
[#permalink]
Show Tags
18 Nov 2015, 03:59
4
gmatser1 wrote:
Can someone explain why the first and last digits are 5^2 and not just 5 options?
since repetitions are allowed, there are 5 ways to choose the letters for the first slot, and five ways to choose for last slot
thats 5 x _ x _ x _ x 5
you still have to choose 1 digit and 2 letters between. For the letters, again, repetitions are allowed so each slot can be filled in 5 ways. For digit, it can only be filled in 3 ways. However, since the digit can be in any one of the 3 middle slots, so we multiply by 3 again:
5 x 3 x 5 x 5 x 5 or 5 x 5 x 3 x 5 x 5 or 5 x 5 x 5 x 3 x 5
Re: A 5-digit code consists of one number digit chosen from 1, 2
[#permalink]
Show Tags
10 May 2017, 11:31
SFF wrote:
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 bit logic is needed for this problem.there are three letters and we need to form 4 letters code. _ _ _ (A,B,or C).For every code we have to repeat one letter.
So, the solution will be : 3*4/2! = 36 . hope it helps.
Re: A 5-digit code consists of one number digit chosen from 1, 2
[#permalink]
Show Tags
12 Jun 2017, 02:22
VeritasPrepKarishma wrote:
samarthgupta wrote:
GyanOne wrote:
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
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 :-
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
The question is not difficult to do once it's understood. I think the question is very badly written, because I had to read it at least thrice to get to know what exactly it is asking. I found the language really bad on this question
Re: A 5-digit code consists of one number digit chosen from 1, 2
[#permalink]
Show Tags
14 Jun 2017, 16:27
1
SFF wrote:
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
Let’s determine the number of ways we can produce each digit.
If L denotes a letter digit and N denotes a number digit, the possibilities for the code are L-N-L-L-L, L-L-N-L-L, and L-L-L-N-L. Note that there are an equal number of possible codes for each of these formats, therefore we will find the number of L-N-L-L-L codes and multiply the result by three.
Since the first digit must be a letter, we have 5 options for the first digit. Since the second digit is a number, there are 3 options for the second digit. For the third, fourth, and fifth digits, we have 5 options each. In total, there are 5 x 3 x 5 x 5 x 5 = 1875 L-N-L-L-L codes. Since the total number of codes is three times that, there are 1875 x 3 = 5625 possible codes.
Re: A 5-digit code consists of one number digit chosen from 1, 2
[#permalink]
Show Tags
28 Jun 2018, 11:41
SFF wrote:
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 is to be formed. Each code has one number from {1,2,3} & 4 letters from {A,B,C,D,E}
The constraint is that first & last place in the code has to be a letter & the items in each place can be repeated. Hence we can have a code as AAA1A or D2DDD or AB3CD, etc.
The code structure can be expressed as (Letter)(Letter)(Letter)(#)(Letter) or (Letter)(Letter)(#)(Letter)(Letter) or (letter)(#)(Letter)(Letter)(Letter)
Re: A 5-digit code consists of one number digit chosen from 1, 2
[#permalink]
Show Tags
14 Apr 2019, 10:56
SFF wrote:
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
5 letter and 3 digits letter takes first and last postion and digits can take 3 positions so 5*5*5*3*5 - 1875 ; 3 postions of digit possible 1875 * 3; 5625 IMO E
gmatclubot
Re: A 5-digit code consists of one number digit chosen from 1, 2
[#permalink]
14 Apr 2019, 10:56
close
50% (02:40) correct 
