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

It is currently 20 Jun 2018, 21:47

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

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

The letters D, G, I, I , and T can be used to form 5-letter strings as

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

Hide Tags

BSchool Forum Moderator
User avatar
V
Joined: 26 Feb 2016
Posts: 2831
Location: India
GPA: 3.12
Premium Member
The letters D, G, I, I , and T can be used to form 5-letter strings as [#permalink]

Show Tags

New post 07 Mar 2018, 11:31
1
Bunuel wrote:
The letters D, G, I, I , and T can be used to form 5-letter strings as DIGIT or DGIIT. Using these letters, how many 5-letter strings can be formed in which the two occurrences of the letter I are separated by at least one other letter?

A) 12
B) 18
C) 24
D) 36
E) 48


Total possibilities of forming 5 letter words with alphabets D I G I T is 5*4*3*2*1

Here, there are 5 possibilities for the first position, 4 for the second position, 3 for the third position,
2 for the second position, and 1 for the final position

However, since we have 2 I's, we divide the total possibilities by 2

Total possibilities with digits D I G I T are \(\frac{5*4*3*2*1}{2} = 60\)

Possibilities when the two I's are together
Consider the 2 I's as 1 unit. Let's call it X.
For the digits X D G T, Total possibilities are 4*3*2*1 = 24

Possibilities where I's are not together = Total possibilities - possibilities when they come together = 60-24 = 36

Therefore, the total possibilities in which the two occurrences of I aren't together is 36(Option D)
_________________

You've got what it takes, but it will take everything you've got

Manager
Manager
avatar
S
Joined: 23 Sep 2016
Posts: 231
Premium Member Reviews Badge
Re: The letters D, G, I, I , and T can be used to form 5-letter strings as [#permalink]

Show Tags

New post 09 Mar 2018, 01:26
pushpitkc wrote:
Bunuel wrote:
The letters D, G, I, I , and T can be used to form 5-letter strings as DIGIT or DGIIT. Using these letters, how many 5-letter strings can be formed in which the two occurrences of the letter I are separated by at least one other letter?

A) 12
B) 18
C) 24
D) 36
E) 48


Total possibilities of forming 5 letter words with alphabets D I G I T is 5*4*3*2*1

Here, there are 5 possibilities for the first position, 4 for the second position, 3 for the third position,
2 for the second position, and 1 for the final position

However, since we have 2 I's, we divide the total possibilities by 2

Total possibilities with digits D I G I T are \(\frac{5*4*3*2*1}{2} = 60\)

Possibilities when the two I's are together
Consider the 2 I's as 1 unit. Let's call it X.
For the digits X D G T, Total possibilities are 4*3*2*1 = 24

Possibilities where I's are not together = Total possibilities - possibilities when they come together = 60-24 = 36

Therefore, the total possibilities in which the two occurrences of I aren't together is 36(Option D)

I solved this by this method No case is possible with 2 I together so if we subtract total no. of cases with the cases of 2 I together than that will be our answer
total cases are 5!/2! (total letters/repeated letters)
together case is 4!*2!/2!(two i together counted as 1 then total letter is 4* both 2 can exchange their places/2 repeated no.)
=4!
5*4*3-4*3*2*1
=4*3(5-2)
=12(3)=36
Expert Post
Director
Director
User avatar
B
Joined: 17 Dec 2012
Posts: 635
Location: India
Re: The letters D, G, I, I , and T can be used to form 5-letter strings as [#permalink]

Show Tags

New post 28 Mar 2018, 22:24
Bunuel wrote:
The letters D, G, I, I , and T can be used to form 5-letter strings as DIGIT or DGIIT. Using these letters, how many 5-letter strings can be formed in which the two occurrences of the letter I are separated by at least one other letter?

A) 12
B) 18
C) 24
D) 36
E) 48

Leftmost arrangement of constraints:I_I_ _

Using formula, we have the number of permutations as (2!/2!) * 3!* 6=36

For explanation of the formula see link below and look for link Permutation with constraints.
_________________

Srinivasan Vaidyaraman
Sravna
http://www.sravnatestprep.com/best-online-gre-preparation.php

Improve Intuition and Your Score
Systematic Approaches

Director
Director
User avatar
G
Joined: 09 Mar 2016
Posts: 605
The letters D, G, I, I , and T can be used to form 5-letter strings as [#permalink]

Show Tags

New post 01 May 2018, 05:04
Bunuel wrote:
The letters D, G, I, I , and T can be used to form 5-letter strings as DIGIT or DGIIT. Using these letters, how many 5-letter strings can be formed in which the two occurrences of the letter I are separated by at least one other letter?

A) 12
B) 18
C) 24
D) 36
E) 48



hello again :) pushpitkc, generis, niks18

Here is my solution, its a bit different unlike others` :) i replaced D,G,I,T, I with following letters \(A\), \(B\), \(C\), \(D\), \(E\), so here, i simply modified question a bit yet arrived at the correct solution.

Using these letters, how many 5-letter strings can be formed in which the two occurrences of the letters A and E are separated by at least one other letter?

So here is my reasoning

\(A\), \(B\), \(C\), \(D\), \(E\)

Step One Fix \(E\), and count number of ways of \(A\) which is 3! * 2 = 12 (we mutiply by 2, because A can be on either left or right side)

Step Two Fix \(A\), and count number of ways of \(E\) which is 3! * 2 = 12 (we mutiply by 2, because E can be on either left or right side)

Step Three Now we have to count number of ways of these three letters \(B\), \(C\), \(D\) which is 3! * 2 = 12 ( mutiply by 2, because ORDER MATTERS/ PERMUTATION)

So total number of ways is \(12+12+12 = 36\)

i doubt that my third step is correct...though ... any ideas ? Quick tips? :) please :-)

pushpitkc, generis niks18 :) did you see this post :? :)
1 KUDOS received
PS Forum Moderator
avatar
D
Joined: 25 Feb 2013
Posts: 1142
Location: India
GPA: 3.82
GMAT ToolKit User Premium Member Reviews Badge
Re: The letters D, G, I, I , and T can be used to form 5-letter strings as [#permalink]

Show Tags

New post 02 May 2018, 09:24
1
dave13 wrote:
Bunuel wrote:
The letters D, G, I, I , and T can be used to form 5-letter strings as DIGIT or DGIIT. Using these letters, how many 5-letter strings can be formed in which the two occurrences of the letter I are separated by at least one other letter?

A) 12
B) 18
C) 24
D) 36
E) 48



hello again :) pushpitkc, generis, niks18

Here is my solution, its a bit different unlike others` :) i replaced D,G,I,T, I with following letters \(A\), \(B\), \(C\), \(D\), \(E\), so here, i simply modified question a bit yet arrived at the correct solution.

Using these letters, how many 5-letter strings can be formed in which the two occurrences of the letters A and E are separated by at least one other letter?

So here is my reasoning

\(A\), \(B\), \(C\), \(D\), \(E\)

Step One Fix \(E\), and count number of ways of \(A\) which is 3! * 2 = 12 (we mutiply by 2, because A can be on either left or right side)

Step Two Fix \(A\), and count number of ways of \(E\) which is 3! * 2 = 12 (we mutiply by 2, because E can be on either left or right side)

Step Three Now we have to count number of ways of these three letters \(B\), \(C\), \(D\) which is 3! * 2 = 12 ( mutiply by 2, because ORDER MATTERS/ PERMUTATION)

So total number of ways is \(12+12+12 = 36\)

i doubt that my third step is correct...though ... any ideas ? Quick tips? :) please :-)

pushpitkc, generis niks18 :) did you see this post :? :)


Hi dave13

I am finding it hard to understand your approach :dazed

However without complicating the matter, the question can be simply solved as

Total Number of Arrangements Possible - Number of arrangements where two I's are always together = Number of ways where two I's are never together.

You simply need to calculate the values of the above equation. Refer to pushpitkc 's solution above for clarity
Intern
Intern
avatar
B
Joined: 18 Apr 2013
Posts: 33
The letters D, G, I, I , and T can be used to form 5-letter strings as [#permalink]

Show Tags

New post 09 May 2018, 10:52
Hi Bunuel, can I check why the no of ways when the Is are placed together is 4!? It can be located _D_G_T_. Why isn't it 4 ways? Thanks in advance!
Expert Post
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 46214
Re: The letters D, G, I, I , and T can be used to form 5-letter strings as [#permalink]

Show Tags

New post 10 May 2018, 01:18
roastedchips wrote:
Hi Bunuel, can I check why the no of ways when the Is are placed together is 4!? It can be located _D_G_T_. Why isn't it 4 ways? Thanks in advance!


In addition to that D, G, and T, could be arranged in 3! ways, so total = 4*3! = 4!.

Or, consider two I's as one unit {II}. We'll have 4 units: {D}, {G}, {T}, and {II}. The number of arrangements is 4!.

Hope it's clear.
_________________

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

Re: The letters D, G, I, I , and T can be used to form 5-letter strings as   [#permalink] 10 May 2018, 01:18

Go to page   Previous    1   2   [ 27 posts ] 

Display posts from previous: Sort by

The letters D, G, I, I , and T can be used to form 5-letter strings as

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


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®.