GMAT Question of the Day - Daily to your Mailbox; hard ones only

It is currently 19 Oct 2018, 12:57

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.

Close

Request Expert Reply

Confirm Cancel

How many four-digit positive integers can be formed by using

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

Hide Tags

Intern
Intern
avatar
Joined: 04 Mar 2012
Posts: 39
How many four-digit positive integers can be formed by using  [#permalink]

Show Tags

New post 22 May 2012, 09:35
7
30
00:00
A
B
C
D
E

Difficulty:

  85% (hard)

Question Stats:

49% (02:06) correct 51% (02:13) wrong based on 359 sessions

HideShow timer Statistics

How many four-digit positive integers can be formed by using the digits from 1 to 9 so that two digits are equal to each other and the remaining two are also equal to each other but different from the other two ?

A. 400
B. 1728
C. 108
D. 216
E. 432

I am getting answer as 432, but the OA is D. Here is the approach I used.

First number can be selected from 1 to 9 - in 9 ways, 2nd number has to be same as the one selected so only 1 way, Third number can be selected from remaining 8 numbers in 8 ways and 4th number again has to be only 1
so in total (9*1*8*1)*4!/2!*2! = 72*6 = 432 Can anyone please explain what mistake I am making here? Thanks!
Most Helpful Expert Reply
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 50002
Re: How many four-digit positive integers can be formed by using  [#permalink]

Show Tags

New post 22 May 2012, 10:35
7
5
gmihir wrote:
How many four-digit positive integers can be formed by using the digits from 1 to 9 so that two digits are equal to each other and the remaining two are also equal to each other but different from the other two ?

A. 400
B. 1728
C. 108
D. 216
E. 432

I am getting answer as 432, but the OA is D. Here is the approach I used.

First number can be selected from 1 to 9 - in 9 ways, 2nd number has to be same as the one selected so only 1 way, Third number can be selected from remaining 8 numbers in 8 ways and 4th number again has to be only 1
so in total (9*1*8*1)*4!/2!*2! = 72*6 = 432 Can anyone please explain what mistake I am making here? Thanks!


XXYY can be arranged in \(\frac{4!}{2!2!}=6\) ways (# of arrangements of 4 letters out of which 2 X's and 2 Y's are identical);
# of ways we can select 2 distinct digits out of 9 is \(C^2_9=36\);

Total # of integers that can be formed is 6*36=216.

Answer: D.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | 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

Most Helpful Community Reply
Intern
Intern
avatar
Joined: 27 Oct 2011
Posts: 12
Schools: Cambridge
Re: How many four-digit positive integers can be formed by using  [#permalink]

Show Tags

New post 22 May 2012, 10:28
5
2
gmihir wrote:
How many four-digit positive integers can be formed by using the digits from 1 to 9 so that two digits are equal to each other and the remaining two are also equal to each other but different from the other two ?

A. 400
B. 1728
C. 108
D. 216
E. 432

I am getting answer as 432, but the OA is D. Here is the approach I used.

First number can be selected from 1 to 9 - in 9 ways, 2nd number has to be same as the one selected so only 1 way, Third number can be selected from remaining 8 numbers in 8 ways and 4th number again has to be only 1
so in total (9*1*8*1)*4!/2!*2! = 72*6 = 432 Can anyone please explain what mistake I am making here? Thanks!




The number of ways u can chose two digits from 9 digits is 9C2=(9*8)/2.
And the no of ways u can arrange two digits like asked is 4C2=(4*3)/2.

total ways is 36*6=216.

Hope that helps.
If you like it give me Kudos :-D
General Discussion
Current Student
avatar
Joined: 21 May 2012
Posts: 93
Location: United States (CA)
Reviews Badge
Re: How many four-digit positive integers can be formed by using  [#permalink]

Show Tags

New post 22 May 2012, 10:29
1
This one is confusing to me. I know how to find the answer in a much more brute force manner.

Think about how many possible combinations you can make with 2 numbers (say 1 and 2)

1122, 1212, 1221, 2121, 2211, 2112 (6 combinations)

The number 1 has 8 numbers it can pair off with
The number 2 has 7 numbers it can pair off with
The number 3 has 6 numbers it can pair off with so on and so forth

\(= (8+7+6+5+4+3+2+1) * 6\)

\(= 36 * 6\)

\(= 216\)
Intern
Intern
avatar
Joined: 27 Oct 2011
Posts: 12
Schools: Cambridge
Re: How many four-digit positive integers can be formed by using  [#permalink]

Show Tags

New post 22 May 2012, 10:43
Cares wrote:
This one is confusing to me. I know how to find the answer in a much more brute force manner.

Think about how many possible combinations you can make with 2 numbers (say 1 and 2)

1122, 1212, 1221, 2121, 2211, 2112 (6 combinations)

The number 1 has 8 numbers it can pair off with
The number 2 has 7 numbers it can pair off with
The number 3 has 6 numbers it can pair off with so on and so forth

\(= (8+7+6+5+4+3+2+1) * 6\)

\(= 36 * 6\)

\(= 216\)


Regarding the pairing you are right.It is the basic way of understanding combinations of numbers.
here you are picking 2 numbers to form a set from 9 available numbers,which turns out to be 9C2.
Basically picking of k numbers from a n set of numbers gives nCk ways of possibilities.

Hope its clear to you now.
Manager
Manager
avatar
Joined: 26 Dec 2011
Posts: 93
Re: How many four-digit positive integers can be formed by using  [#permalink]

Show Tags

New post 24 May 2012, 02:09
I understood the solution of Bunuel, but can you explain where did gmihir went wrong... Thanks in advance.
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 50002
Re: How many four-digit positive integers can be formed by using  [#permalink]

Show Tags

New post 24 May 2012, 02:16
1
1
pavanpuneet wrote:
I understood the solution of Bunuel, but can you explain where did gmihir went wrong... Thanks in advance.


In that solution the # of ways to choose 2 different numbers for X and Y is 9*8=72, but it should be \(C^2_9=36\), so gmihir's solution counts the same numbers twice.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | 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

Senior Manager
Senior Manager
avatar
B
Joined: 24 Aug 2009
Posts: 475
Schools: Harvard, Columbia, Stern, Booth, LSB,
Re: How many four-digit positive integers can be formed by using  [#permalink]

Show Tags

New post 16 Sep 2012, 06:09
Bunuel wrote:
gmihir wrote:
How many four-digit positive integers can be formed by using the digits from 1 to 9 so that two digits are equal to each other and the remaining two are also equal to each other but different from the other two ?

A. 400
B. 1728
C. 108
D. 216
E. 432

I am getting answer as 432, but the OA is D. Here is the approach I used.

First number can be selected from 1 to 9 - in 9 ways, 2nd number has to be same as the one selected so only 1 way, Third number can be selected from remaining 8 numbers in 8 ways and 4th number again has to be only 1
so in total (9*1*8*1)*4!/2!*2! = 72*6 = 432 Can anyone please explain what mistake I am making here? Thanks!


XXYY can be arranged in \(\frac{4!}{2!2!}=6\) ways (# of arrangements of 4 letters out of which 2 X's and 2 Y's are identical);
# of ways we can select 2 distinct digits out of 9 is \(C^2_9=36\);

Total # of integers that can be formed is 6*36=216.

Answer: D.


Hi Bunuel,

I Want to know if the statement changes from "How many four-digit positive integers can be formed by using the digits from 1 to 9" to How many four-digit positive integers can be formed by using the digits from 0 to 9", what would be answer.

As per me answer should be 243. Kindly conform it.
_________________

If you like my Question/Explanation or the contribution, Kindly appreciate by pressing KUDOS.
Kudos always maximizes GMATCLUB worth
-Game Theory

If you have any question regarding my post, kindly pm me or else I won't be able to reply

Director
Director
User avatar
Joined: 22 Mar 2011
Posts: 601
WE: Science (Education)
Re: How many four-digit positive integers can be formed by using  [#permalink]

Show Tags

New post 16 Sep 2012, 06:46
fameatop wrote:
Bunuel wrote:
gmihir wrote:
How many four-digit positive integers can be formed by using the digits from 1 to 9 so that two digits are equal to each other and the remaining two are also equal to each other but different from the other two ?

A. 400
B. 1728
C. 108
D. 216
E. 432

I am getting answer as 432, but the OA is D. Here is the approach I used.

First number can be selected from 1 to 9 - in 9 ways, 2nd number has to be same as the one selected so only 1 way, Third number can be selected from remaining 8 numbers in 8 ways and 4th number again has to be only 1
so in total (9*1*8*1)*4!/2!*2! = 72*6 = 432 Can anyone please explain what mistake I am making here? Thanks!


XXYY can be arranged in \(\frac{4!}{2!2!}=6\) ways (# of arrangements of 4 letters out of which 2 X's and 2 Y's are identical);
# of ways we can select 2 distinct digits out of 9 is \(C^2_9=36\);

Total # of integers that can be formed is 6*36=216.

Answer: D.


Hi Bunuel,

I Want to know if the statement changes from "How many four-digit positive integers can be formed by using the digits from 1 to 9" to How many four-digit positive integers can be formed by using the digits from 0 to 9", what would be answer.

As per me answer should be 243. Kindly conform it.


Yes, the answer should be 243.

Assume the first digit is A, for which we have 9 choices, because it cannot be 0. Then, we have 3 choices where to place the second digit A.
Now, for the other different digit B, we have again 9 choices (different from A but can be 0), and the places of the two B's are uniquely determined.
Therefore, total number of possibilities is 9*3*9 = 243.

A similar logic can be applied also for the original question:
9*3*8 = 216
9 possibilities for A
3 places to choose from for the second A
8 possibilites for B (cannot be A and cannot be 0), no places to choose, already determined
_________________

PhD in Applied Mathematics
Love GMAT Quant questions and running.

Manager
Manager
avatar
Joined: 25 Jun 2012
Posts: 62
Location: India
WE: General Management (Energy and Utilities)
Re: How many four-digit positive integers can be formed by using  [#permalink]

Show Tags

New post 18 Sep 2012, 23:07
we units 1,2,3,4,5,6,7,8,9
total 9 digits
Now we want arragement xxyy

x - can be selected in 9 ways
x - same as first x - 1 way
y - can be selected from remaining 8 digits - 8 ways
y - same as first y - 1 way

so total# of ways = 9*1*8*1 = 72
not the 4 digits can be organised in 4C2 ways = 6 ways

So total arragements = 72*6 = 432

Am I wrong in thinking? & I didnt understand the concept mentioned by Bunuel here...

P & C is my weakest point in Quant :oops:
Director
Director
User avatar
Joined: 22 Mar 2011
Posts: 601
WE: Science (Education)
Re: How many four-digit positive integers can be formed by using  [#permalink]

Show Tags

New post 19 Sep 2012, 00:00
2
2
bhavinshah5685 wrote:
we units 1,2,3,4,5,6,7,8,9
total 9 digits
Now we want arragement xxyy

x - can be selected in 9 ways
x - same as first x - 1 way
y - can be selected from remaining 8 digits - 8 ways
y - same as first y - 1 way

so total# of ways = 9*1*8*1 = 72
not the 4 digits can be organised in 4C2 ways = 6 ways

So total arragements = 72*6 = 432

Am I wrong in thinking? & I didnt understand the concept mentioned by Bunuel here...

P & C is my weakest point in Quant :oops:



the 4 digits can be organised in 4C2 ways = 6 ways
You have to divide by 2, so only 3 ways.
Or, 9 possibilities for the first digit, this can be placed for the second time in 3 places, then in the remaining places have another different digit - 8 possibilities.
A total of 9*3*8=216.

4C2 would work if you choose two distinct digit out of 9, which is 9C2 = 36 and you don't care about the order in which you have chosen them, take 2 of each type, then arrange them, which is 4C2 = 6.
This will give you again 9C2*4C2 = 36*6 = 216.

You cannot in one stage take order into account, then in the next stage ignore order. Choosing two digits as 9*8 means you distinguish between which one is chosen first. For example, once you considered x = 4 and y = 1, which you can arrange in 4C2 = 6 ways. But in this 6 arrangements are also include those with a 1 in the first place.
Then you considered the pair x = 1 and y = 4, for which again 4C2 = 6 arrangements, but these are identical to the previous ones. 4C2 counts arrangements without distinguishing whether you first place x or y.
_________________

PhD in Applied Mathematics
Love GMAT Quant questions and running.

Senior Manager
Senior Manager
User avatar
B
Joined: 03 Sep 2012
Posts: 382
Location: United States
Concentration: Healthcare, Strategy
GMAT 1: 730 Q48 V42
GPA: 3.88
WE: Medicine and Health (Health Care)
Re: How many four-digit positive integers can be formed by using  [#permalink]

Show Tags

New post 24 Sep 2012, 03:42
Numbers are 1, 2, 3, 4, 5, 6, 7, 8, 9 ie 9 different numbers

Lets assume we have a XXYY scenario , this means that we could have 9x1x8x1 = 72 different numbers

We could also have XyXy combination so that is also 72 different numbers

We could have a XYYX combination and that is 72 more different numbers ...

72 x 3 = 216...

D is the answer ...
_________________

"When you want to succeed as bad as you want to breathe, then you’ll be successful.” - Eric Thomas

Manager
Manager
User avatar
B
Joined: 16 Mar 2016
Posts: 129
Location: France
GMAT 1: 660 Q47 V33
GPA: 3.25
GMAT ToolKit User
How many four-digit positive integers can be formed by using  [#permalink]

Show Tags

New post 05 May 2016, 05:27
Stage 1 : Number of ways to select 2 different digits form 9 => (9 * 8 )/2 =36

Stage 2 : Number of ways to arrange AABB (MISSISSIPPI rule) = (4 * 3 *2) / (2*2) = 6

Fondamental Counting Principle = 36 * 6 = 216
Manager
Manager
User avatar
Joined: 18 May 2016
Posts: 67
Concentration: Finance, International Business
GMAT 1: 720 Q49 V39
GPA: 3.7
WE: Analyst (Investment Banking)
How many four-digit positive integers can be formed by using  [#permalink]

Show Tags

New post Updated on: 01 Jun 2016, 12:33
surbhi87 wrote:
yeah i did but don't quite understand the logic :( could you pls elaborate a little?


If anyone else wanders into this forum and has the same question like Surbhi87 and I:

I thought perhaps, the simple explanation for why the number of ways to pick to digits by simply multiplying 9 x 8 = 72 is because this is not a permutation, but a combination problem, where the order of selecting numbers does not matter. Am I correct?
_________________

Please kindly +Kudos if my posts or questions help you!

My debrief: Self-study: How to improve from 620(Q39,V36) to 720(Q49,V39) in 25 days!


Originally posted by fantaisie on 01 Jun 2016, 08:30.
Last edited by fantaisie on 01 Jun 2016, 12:33, edited 1 time in total.
Manager
Manager
User avatar
B
Joined: 16 Mar 2016
Posts: 129
Location: France
GMAT 1: 660 Q47 V33
GPA: 3.25
GMAT ToolKit User
Re: How many four-digit positive integers can be formed by using  [#permalink]

Show Tags

New post 01 Jun 2016, 09:21
fantaisie wrote:
surbhi87 wrote:
yeah i did but don't quite understand the logic :( could you pls elaborate a little?


If anyone else wanders into this forum and has the same question like Surbhi87 and I:

I though perhaps, the simple explanation for why the number of ways to pick to digits by simply multiplying 9 x 8 = 72 is because this is not a permutation, but a combination problem, where the order of selecting numbers does not matter. Am I correct?


Exactly, you can have 1144, or 4141, or 4411, or etc...
Since you have two pairs of equal digits, the order doesn't matter here.
Manager
Manager
avatar
Joined: 01 Mar 2014
Posts: 117
Schools: Tepper '18
Re: How many four-digit positive integers can be formed by using  [#permalink]

Show Tags

New post 04 Jun 2016, 12:24
Bunuel wrote:
pavanpuneet wrote:
I understood the solution of Bunuel, but can you explain where did gmihir went wrong... Thanks in advance.


In that solution the # of ways to choose 2 different numbers for X and Y is 9*8=72, but it should be \(C^2_9=36\), so gmihir's solution counts the same numbers twice.


I always tend to miss the cases where a solution is counted twice. Could you please explain as to how i can get it right?
Target Test Prep Representative
User avatar
G
Status: Head GMAT Instructor
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 2830
Re: How many four-digit positive integers can be formed by using  [#permalink]

Show Tags

New post 26 Jun 2017, 17:54
3
gmihir wrote:
How many four-digit positive integers can be formed by using the digits from 1 to 9 so that two digits are equal to each other and the remaining two are also equal to each other but different from the other two ?

A. 400
B. 1728
C. 108
D. 216
E. 432


Let’s first find the number of ways we can form a number AABB where A and B denote distinct digits.

For the first digit, we have 9 possible options. If the second digit matches that of the first, then we have 1 option. Then for digit 3, we have 8 possible options. Since digit 4 matches digit 3, we have 1 option.

So, for this scenario, we have 9 x 1 x 8 x 1 = 72 options.

Since we are essentially being asked how many ways we can arrange As and Bs in A-A-B-B, where A and B are numbers 1 through 9, we see that A-A-B-B can be arranged in 4!/(2! x 2!) = 24/4 = 6 ways.

Finally, we should take into account that we counted each number twice when we made this calculation (for instance, when A = 1 and B = 2, AABB = 1122, but when A = 2 and B = 1, BBAA = 1122); therefore, we should divide the result by 2.

Thus, the total number of ways to create a 4-digit number with 2 unique digits is (6 x 72)/2 = 216.

Answer: D
_________________

Jeffery Miller
Head of GMAT Instruction

GMAT Quant Self-Study Course
500+ lessons 3000+ practice problems 800+ HD solutions

Intern
Intern
avatar
Joined: 24 May 2017
Posts: 2
Re: How many four-digit positive integers can be formed by using  [#permalink]

Show Tags

New post 06 Jul 2017, 20:11
Of the three-digit positive integers that have no digits equal to zero, how many have two digits that are equal to each other and the remaining digit different from the other two?

Then how come this question is not 9C2 * 3C2?
Intern
Intern
avatar
B
Joined: 30 Apr 2017
Posts: 14
Re: How many four-digit positive integers can be formed by using  [#permalink]

Show Tags

New post 16 Jul 2017, 04:09
Quote:
the 4 digits can be organised in 4C2 ways = 6 ways
You have to divide by 2, so only 3 ways.
Or, 9 possibilities for the first digit, this can be placed for the second time in 3 places, then in the remaining places have another different digit - 8 possibilities.
A total of 9*3*8=216.

4C2 would work if you choose two distinct digit out of 9, which is 9C2 = 36 and you don't care about the order in which you have chosen them, take 2 of each type, then arrange them, which is 4C2 = 6.
This will give you again 9C2*4C2 = 36*6 = 216.

You cannot in one stage take order into account, then in the next stage ignore order. Choosing two digits as 9*8 means you distinguish between which one is chosen first. For example, once you considered x = 4 and y = 1, which you can arrange in 4C2 = 6 ways. But in this 6 arrangements are also include those with a 1 in the first place.
Then you considered the pair x = 1 and y = 4, for which again 4C2 = 6 arrangements, but these are identical to the previous ones. 4C2 counts arrangements without distinguishing whether you first place x or y.


This is the best explanation of the conundrum by far.
I was also confused at first, not understanding why the answer is 216, not 432.

In fact, if we pick the 2 numbers as a first step using "9 x 8 = 72", we are picking for example 8 & 9 as well as 9 & 8. Which means we have already "arranged" the 2 numbers one time. And then when we arrange the 2 numbers using 4!/2!*2!, we are again arranging the same numbers, now for a second time. That is why double counting is involved.

The way to solve this is to NOT arrange the numbers when we pick them in the first place, meaning we should use "combination" instead of "permutation". So, choose 2 numbers out of 9 (9C2 = 36) and then arrange those 2 numbers in 4!/2!*2! ways.

Hope this helps
Intern
Intern
avatar
B
Joined: 12 Jun 2018
Posts: 5
How many four-digit positive integers can be formed by using  [#permalink]

Show Tags

New post 20 Jul 2018, 10:50
If I arrange two distinct things in four places - 4!/2!*2! = 6
Select two from nine - 9c2
So total - 9*8*6/2 = 216
GMAT Club Bot
How many four-digit positive integers can be formed by using &nbs [#permalink] 20 Jul 2018, 10:50

Go to page    1   2    Next  [ 21 posts ] 

Display posts from previous: Sort by

How many four-digit positive integers can be formed by using

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


Copyright

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

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne

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®.