Find all School-related info fast with the new School-Specific MBA Forum

It is currently 29 Sep 2016, 08:34
GMAT Club Tests

Close

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

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.

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

A 5-digit code consists of one number digit chosen from 1, 2

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

1 KUDOS received
Intern
Intern
avatar
Joined: 09 Sep 2011
Posts: 3
Location: India
Concentration: General Management
Schools: MIT-LGO '17 (S)
GMAT 1: 720 Q49 V38
Followers: 0

Kudos [?]: 24 [1] , given: 1

A 5-digit code consists of one number digit chosen from 1, 2 [#permalink]

Show Tags

New post 08 May 2012, 23:30
1
This post received
KUDOS
20
This post was
BOOKMARKED
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

48% (03:05) correct 52% (01:55) wrong based on 507 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?

A. 375
B. 625
C. 1,875
D. 3,750
E. 5,625
[Reveal] Spoiler: OA
4 KUDOS received
VP
VP
User avatar
Status: Top MBA Admissions Consultant
Joined: 24 Jul 2011
Posts: 1067
GMAT 1: 780 Q51 V48
GRE 1: 1540 Q800 V740
Followers: 112

Kudos [?]: 495 [4] , given: 18

Re: Maths Question on Combinations [#permalink]

Show Tags

New post 08 May 2012, 23:46
4
This post received
KUDOS
1
This post was
BOOKMARKED
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 | Top MBA Rankings and MBA Admissions Blog

Top MBA Admissions Consulting | Top MiM Admissions Consulting

Premium MBA Essay Review|Best MBA Interview Preparation|Exclusive GMAT coaching

Get a FREE Detailed MBA Profile Evaluation | Call us now +91 98998 31738


Last edited by GyanOne on 09 May 2012, 02:58, edited 2 times in total.
Intern
Intern
avatar
Joined: 09 Sep 2011
Posts: 3
Location: India
Concentration: General Management
Schools: MIT-LGO '17 (S)
GMAT 1: 720 Q49 V38
Followers: 0

Kudos [?]: 24 [0], given: 1

Re: Maths Question on Combinations [#permalink]

Show Tags

New post 09 May 2012, 01:14
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 :-

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.
Expert Post
3 KUDOS received
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 34899
Followers: 6498

Kudos [?]: 83030 [3] , given: 10134

Re: A 5-digit code consists of one number digit chosen from 1, 2 [#permalink]

Show Tags

New post 09 May 2012, 01:23
3
This post received
KUDOS
Expert's post
7
This post was
BOOKMARKED
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).

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.
_________________

New to the Math Forum?
Please read this: All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Expert Post
5 KUDOS received
Veritas Prep GMAT Instructor
User avatar
Joined: 16 Oct 2010
Posts: 6921
Location: Pune, India
Followers: 1994

Kudos [?]: 12411 [5] , given: 221

Re: Maths Question on Combinations [#permalink]

Show Tags

New post 09 May 2012, 09:55
5
This post received
KUDOS
Expert's post
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 :-

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
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for $199

Veritas Prep Reviews

Expert Post
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 34899
Followers: 6498

Kudos [?]: 83030 [0], given: 10134

Re: A 5-digit code consists of one number digit chosen from 1, 2 [#permalink]

Show Tags

New post 01 Jul 2013, 00:59
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE

Theory on Combinations: math-combinatorics-87345.html

DS questions on Combinations: search.php?search_id=tag&tag_id=31
PS questions on Combinations: search.php?search_id=tag&tag_id=52

Tough and tricky questions on Combinations: hardest-area-questions-probability-and-combinations-101361.html

_________________

New to the Math Forum?
Please read this: All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Director
Director
User avatar
Status: Verbal Forum Moderator
Joined: 17 Apr 2013
Posts: 635
Location: India
GMAT 1: 710 Q50 V36
GMAT 2: 750 Q51 V41
GMAT 3: 790 Q51 V49
GPA: 3.3
Followers: 56

Kudos [?]: 354 [0], given: 297

Re: Maths Question on Combinations [#permalink]

Show Tags

New post 18 Sep 2013, 05:56
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 :-

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



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
_________________

Like my post Send me a Kudos :) It is a Good manner.
My Debrief: http://gmatclub.com/forum/how-to-score-750-and-750-i-moved-from-710-to-189016.html

Expert Post
Veritas Prep GMAT Instructor
User avatar
Joined: 16 Oct 2010
Posts: 6921
Location: Pune, India
Followers: 1994

Kudos [?]: 12411 [0], given: 221

Re: Maths Question on Combinations [#permalink]

Show Tags

New post 18 Sep 2013, 07:02
honchos wrote:
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
My Blog

Get started with Veritas Prep GMAT On Demand for $199

Veritas Prep Reviews

GMAT Club Legend
GMAT Club Legend
User avatar
Joined: 09 Sep 2013
Posts: 11740
Followers: 529

Kudos [?]: 149 [0], given: 0

Premium Member
Re: A 5-digit code consists of one number digit chosen from 1, 2 [#permalink]

Show Tags

New post 05 Nov 2014, 07:58
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

GMAT Books | GMAT Club Tests | Best Prices on GMAT Courses | GMAT Mobile App | Math Resources | Verbal Resources

Intern
Intern
avatar
Joined: 30 Oct 2014
Posts: 1
Followers: 0

Kudos [?]: 0 [0], given: 9

Re: A 5-digit code consists of one number digit chosen from 1, 2 [#permalink]

Show Tags

New post 12 May 2015, 05:36
1
This post was
BOOKMARKED
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!

=5*3*5*5*5 + 5*5*3*5*5 + 5*5*5*3*5
=5*3*5*5*5*3
=5625.
1 KUDOS received
Manager
Manager
avatar
Joined: 08 May 2015
Posts: 105
GMAT 1: 630 Q39 V38
GMAT 2: 670 Q44 V38
GMAT 3: 750 Q49 V44
Followers: 2

Kudos [?]: 58 [1] , given: 14

Re: A 5-digit code consists of one number digit chosen from 1, 2 [#permalink]

Show Tags

New post 17 Jun 2015, 18:25
1
This post received
KUDOS
Since it will have "3*3 in the number", it will have to be divisible by 9. The only choice is E.
Manager
Manager
avatar
Joined: 07 Feb 2015
Posts: 84
Followers: 1

Kudos [?]: 10 [0], given: 28

Re: A 5-digit code consists of one number digit chosen from 1, 2 [#permalink]

Show Tags

New post 18 Jun 2015, 06:56
Can someone explain why the first and last digits are 5^2 and not just 5 options?
1 KUDOS received
Intern
Intern
avatar
Joined: 13 Sep 2015
Posts: 22
Followers: 0

Kudos [?]: 6 [1] , given: 239

Re: A 5-digit code consists of one number digit chosen from 1, 2 [#permalink]

Show Tags

New post 18 Nov 2015, 03:59
1
This post received
KUDOS
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

so 5 x 5 x 5 x 5 x 3 x 3 = 5^4 x 3^2 = 5625.
Intern
Intern
avatar
Joined: 08 Dec 2015
Posts: 30
Followers: 1

Kudos [?]: 1 [0], given: 57

Re: A 5-digit code consists of one number digit chosen from 1, 2 [#permalink]

Show Tags

New post 14 Dec 2015, 15:00
Hey guys, I solved this similarly, but it took me some time (4.5mins). Can someone explain what I can/should do differently? Here are my steps below.

1. Deep breath (bc combo problems give me anxiety)

2. 5 letter word ( _ _ _ _ _ )

3. First and last choices are letters ( 5 _ _ _ 5)

4. The middle includes a number so the choices are:
5 3 5 5 5
5 5 3 5 5
5 5 5 3 5

5. If we add these up, we get:
3 * (5*3*5*5*5) = 5^4 * 3^2 = 625*9 = something slightly less than 6250... (I'm quickly running out of steam here...)

6. Choose E because it is the closest answer and I've already spent way too much time on this problem
Intern
Intern
avatar
Joined: 28 Dec 2015
Posts: 40
Followers: 1

Kudos [?]: 1 [0], given: 61

CAT Tests
Re: A 5-digit code consists of one number digit chosen from 1, 2 [#permalink]

Show Tags

New post 13 Jul 2016, 21:21
_ _ _ _ _

The first and last letter must be one of the letter A,B,C,D and E.

one digit would be a number out of 1,2,and 3.

So,we have 5*3*5*5*5-Digit takes the second position-1875

If digit- takes third position we have 5*5*3*5*5-1875

If digit takes the fourth position,we have 5*5*5*3*5-1875

All add up to 5625
Re: A 5-digit code consists of one number digit chosen from 1, 2   [#permalink] 13 Jul 2016, 21:21
    Similar topics Author Replies Last post
Similar
Topics:
2 Experts publish their posts in the topic We make 4 digit codes and each digit of the code form from 1, 2, 3, an MathRevolution 3 26 Jan 2016, 22:07
4 Experts publish their posts in the topic How many 5 digit nos are there if the 2 leftmost digits are abhi47 5 02 Feb 2012, 04:59
2 Experts publish their posts in the topic A security company can use number 1 -7 to create a 5 digit TheSituation 5 23 Feb 2010, 16:39
17 Experts publish their posts in the topic In a 5 digit ID number, what is the probability of exactly t Bunuel 27 30 Dec 2009, 17:08
3 Experts publish their posts in the topic How many 5 digit ( digit = 0 - 9 ) zip codes can exist in wh HopefulOldie 10 10 Nov 2009, 10:47
Display posts from previous: Sort by

A 5-digit code consists of one number digit chosen from 1, 2

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  


GMAT Club MBA Forum Home| About| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group and phpBB SEO

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.