It is currently 20 Nov 2017, 10:50

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

A researcher plans to identify each participant in a certain

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

Hide Tags

Intern
Intern
avatar
Joined: 20 Nov 2013
Posts: 27

Kudos [?]: 14 [0], given: 189

Schools: LBS '17
Re: A researcher plans to identify each participant in a certain [#permalink]

Show Tags

New post 25 Feb 2014, 08:57
Bunuel wrote:
sarb wrote:
A researcher plans to identify each participant in a certain medical experiment with a code consisting of either a single letter or a pair of distinct letters written in alphabetical order. What is the least number of letters that can be used if there are 12 participants, and each participant is to receive a different code?

A. 4
B. 5
C. 6
D. 7
E. 8


Say there are minimum of \(n\) letters needed, then;

The # of single letter codes possible would be \(n\) itself;
The # of pair of distinct letters codes possible would be \(C^2_n\) (in alphabetical order);

We want \(C^2_n+n\geq{12}\) --> \(\frac{n(n-1)}{2}+n\geq{12}\) --> \(n(n-1)+2n\geq{24}\) --> \(n(n+1)\geq{24}\) --> \(n_{min}=5\).

Answer: B.

Hope it's clear.



I have a questions here:
How did we get from \(n(n+1)\geq{24}\) to \(n_{min}=5\)

Kudos [?]: 14 [0], given: 189

Expert Post
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 42263

Kudos [?]: 132792 [0], given: 12373

Re: A researcher plans to identify each participant in a certain [#permalink]

Show Tags

New post 25 Feb 2014, 09:39
amz14 wrote:
Bunuel wrote:
sarb wrote:
A researcher plans to identify each participant in a certain medical experiment with a code consisting of either a single letter or a pair of distinct letters written in alphabetical order. What is the least number of letters that can be used if there are 12 participants, and each participant is to receive a different code?

A. 4
B. 5
C. 6
D. 7
E. 8


Say there are minimum of \(n\) letters needed, then;

The # of single letter codes possible would be \(n\) itself;
The # of pair of distinct letters codes possible would be \(C^2_n\) (in alphabetical order);

We want \(C^2_n+n\geq{12}\) --> \(\frac{n(n-1)}{2}+n\geq{12}\) --> \(n(n-1)+2n\geq{24}\) --> \(n(n+1)\geq{24}\) --> \(n_{min}=5\).

Answer: B.

Hope it's clear.



I have a questions here:
How did we get from \(n(n+1)\geq{24}\) to \(n_{min}=5\)


By trial and error:
If n=4, then n(n+1)=20<24;
If n=5, then n(n+1)=30>24.

Hence, \(n_{min}=5\).

Try similar questions to practice: a-researcher-plans-to-identify-each-participant-in-a-certain-134584.html#p1296049

Hope this helps.
_________________

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

Kudos [?]: 132792 [0], given: 12373

2 KUDOS received
Intern
Intern
avatar
Joined: 16 Feb 2014
Posts: 1

Kudos [?]: 2 [2], given: 0

Re: A researcher plans to identify each participant in a certain [#permalink]

Show Tags

New post 13 Mar 2014, 15:03
2
This post received
KUDOS
Bunuel wrote:
sarb wrote:
A researcher plans to identify each participant in a certain medical experiment with a code consisting of either a single letter or a pair of distinct letters written in alphabetical order. What is the least number of letters that can be used if there are 12 participants, and each participant is to receive a different code?

A. 4
B. 5
C. 6
D. 7
E. 8


Say there are minimum of \(n\) letters needed, then;

The # of single letter codes possible would be \(n\) itself;
The # of pair of distinct letters codes possible would be \(C^2_n\) (in alphabetical order);

We want \(C^2_n+n\geq{12}\) --> \(\frac{n(n-1)}{2}+n\geq{12}\) --> \(n(n-1)+2n\geq{24}\) --> \(n(n+1)\geq{24}\) --> \(n_{min}=5\).

Answer: B.

Hope it's clear.


I am still having problems with this question. Why do we devide the combinations formula into n(n-1)/2?
Shouldnt it be 2!/n!(2-n)! ?

Kudos [?]: 2 [2], given: 0

Expert Post
3 KUDOS received
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 42263

Kudos [?]: 132792 [3], given: 12373

Re: A researcher plans to identify each participant in a certain [#permalink]

Show Tags

New post 14 Mar 2014, 02:48
3
This post received
KUDOS
Expert's post
4
This post was
BOOKMARKED
RebekaMo wrote:
Bunuel wrote:
sarb wrote:
A researcher plans to identify each participant in a certain medical experiment with a code consisting of either a single letter or a pair of distinct letters written in alphabetical order. What is the least number of letters that can be used if there are 12 participants, and each participant is to receive a different code?

A. 4
B. 5
C. 6
D. 7
E. 8


Say there are minimum of \(n\) letters needed, then;

The # of single letter codes possible would be \(n\) itself;
The # of pair of distinct letters codes possible would be \(C^2_n\) (in alphabetical order);

We want \(C^2_n+n\geq{12}\) --> \(\frac{n(n-1)}{2}+n\geq{12}\) --> \(n(n-1)+2n\geq{24}\) --> \(n(n+1)\geq{24}\) --> \(n_{min}=5\).

Answer: B.

Hope it's clear.


I am still having problems with this question. Why do we devide the combinations formula into n(n-1)/2?
Shouldnt it be 2!/n!(2-n)! ?


\(C^2_n=\frac{n!}{2!(n-2)!}\). Now, notice that \(n!=(n-2)!*(n-1)*n\), hence \(C^2_n=\frac{n!}{2!(n-2)!}=\frac{(n-2)!*(n-1)*n}{2!(n-2)!}=\frac{(n-1)n}{2}\).

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

Kudos [?]: 132792 [3], given: 12373

Senior Manager
Senior Manager
avatar
Joined: 15 Aug 2013
Posts: 301

Kudos [?]: 83 [0], given: 23

Re: A researcher plans to identify each participant in a certain [#permalink]

Show Tags

New post 06 Apr 2014, 12:38
Bunuel wrote:
sarb wrote:
A researcher plans to identify each participant in a certain medical experiment with a code consisting of either a single letter or a pair of distinct letters written in alphabetical order. What is the least number of letters that can be used if there are 12 participants, and each participant is to receive a different code?

A. 4
B. 5
C. 6
D. 7
E. 8


Say there are minimum of \(n\) letters needed, then;

The # of single letter codes possible would be \(n\) itself;
The # of pair of distinct letters codes possible would be \(C^2_n\) (in alphabetical order);

We want \(C^2_n+n\geq{12}\) --> \(\frac{n(n-1)}{2}+n\geq{12}\) --> \(n(n-1)+2n\geq{24}\) --> \(n(n+1)\geq{24}\) --> \(n_{min}=5\).

Answer: B.

Hope it's clear.


Probability and Combinatorics are my weakest subjects by far, so please ignore the rudimentary question.

When we say \(C^2_n+n\geq{12}\) that means that we are going to find a combination of 2 letters out of a group of n letters which in turn would yield "x" amount of options. Correct? If so, why are we adding the n following that equation and more importantly, how does that equation yield 5? When I factor it out, i get n(n+1) >= 24. That yields -1 and 0. Why am I so off here?

My question would be - what does this formula mean and how do you solve it? \(C^2_n+n\geq{12}\)

Also, the question is saying that they need to be in alphabetical order, doesn't that mean that order DOES matter? How does that affect the above equation.

P.S: For what it's worth, I've read the Combinatorics and Probability strategy guide(Manhattan Gmat) and understand the content of the guide but these two topics still elude me. I'm open to learning from another venue if helpful?

EDIT: Simplifying my question.

Kudos [?]: 83 [0], given: 23

Expert Post
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 42263

Kudos [?]: 132792 [0], given: 12373

Re: A researcher plans to identify each participant in a certain [#permalink]

Show Tags

New post 06 Apr 2014, 13:09
russ9 wrote:
Bunuel wrote:
sarb wrote:
A researcher plans to identify each participant in a certain medical experiment with a code consisting of either a single letter or a pair of distinct letters written in alphabetical order. What is the least number of letters that can be used if there are 12 participants, and each participant is to receive a different code?

A. 4
B. 5
C. 6
D. 7
E. 8


Say there are minimum of \(n\) letters needed, then;

The # of single letter codes possible would be \(n\) itself;
The # of pair of distinct letters codes possible would be \(C^2_n\) (in alphabetical order);

We want \(C^2_n+n\geq{12}\) --> \(\frac{n(n-1)}{2}+n\geq{12}\) --> \(n(n-1)+2n\geq{24}\) --> \(n(n+1)\geq{24}\) --> \(n_{min}=5\).

Answer: B.

Hope it's clear.


Probability and Combinatorics are my weakest subjects by far, so please ignore the rudimentary question.

When we say \(C^2_n+n\geq{12}\) that means that a combination of 2 letters out of a group of n letters should yield "x" amount of options. Correct? If so, why are we adding the n following that equation and more importantly, how does that equation yield 5? When I factor it out, i get n(n+1) >= 24. That yields -1 and 0. Why am I so off here?

Also, the question is saying that they need to be in alphabetical order, doesn't that mean that order DOES matter? How does that affect the above equation.

P.S: For what it's worth, I've read the Combinatorics and Probability strategy guide and understand the content of the guide but these two topics still elude me. I'm open to learning from another venue if helpful?


The first advice would be, and I cannot stress this enough, to read the whole thread and follow the links to similar problems.

As for your questions:

Why we are adding n.

The question says that the code can consists of 1 or 2 letters. Now, if we have n letters how many codes we can make?

The # of single letter codes possible would be n itself;
The # of pair of distinct letters codes possible would be \(C^2_n\).

So, out of n letters we can make \(n+C^2_n\) codes: n one-letter codes and \(C^2_n\) two-letter codes.

How the equation yields 5

By trial and error:
If n=4, then n(n+1)=20<24;
If n=5, then n(n+1)=30>24.

Hence, \(n_{min}=5\).

Notice that we have \(n(n+1)\geq{24}\) NOT \(n(n+1)\geq{0}\).

About the alphabetical order.

Check here: a-researcher-plans-to-identify-each-participant-in-a-certain-134584.html#p1150091 and here: a-researcher-plans-to-identify-each-participant-in-a-certain-134584.html#p1296053

Similar questions to practice:
each-student-at-a-certain-university-is-given-a-four-charact-151945.html
all-of-the-stocks-on-the-over-the-counter-market-are-126630.html
if-a-code-word-is-defined-to-be-a-sequence-of-different-126652.html
a-4-letter-code-word-consists-of-letters-a-b-and-c-if-the-59065.html
a-5-digit-code-consists-of-one-number-digit-chosen-from-132263.html
a-company-that-ships-boxes-to-a-total-of-12-distribution-95946.html
a-company-plans-to-assign-identification-numbers-to-its-empl-69248.html
the-security-gate-at-a-storage-facility-requires-a-five-109932.html
all-of-the-bonds-on-a-certain-exchange-are-designated-by-a-150820.html
a-local-bank-that-has-15-branches-uses-a-two-digit-code-to-98109.html
a-researcher-plans-to-identify-each-participant-in-a-certain-134584.html
baker-s-dozen-128782-20.html#p1057502
in-a-certain-appliance-store-each-model-of-television-is-136646.html
m04q29-color-coding-70074.html
john-has-12-clients-and-he-wants-to-use-color-coding-to-iden-107307.html
how-many-4-digit-even-numbers-do-not-use-any-digit-more-than-101874.html
a-certain-stock-exchange-designates-each-stock-with-a-85831.html
the-simplastic-language-has-only-2-unique-values-and-105845.html
m04q29-color-coding-70074.html

Hope this helps.
_________________

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

Kudos [?]: 132792 [0], given: 12373

Intern
Intern
avatar
Joined: 24 Sep 2013
Posts: 3

Kudos [?]: [0], given: 4

Reviews Badge
Re: A researcher plans to identify each participant in a certain [#permalink]

Show Tags

New post 18 May 2014, 00:24
Bunuel wrote:
sarb wrote:
A researcher plans to identify each participant in a certain medical experiment with a code consisting of either a single letter or a pair of distinct letters written in alphabetical order. What is the least number of letters that can be used if there are 12 participants, and each participant is to receive a different code?

A. 4
B. 5
C. 6
D. 7
E. 8


Say there are minimum of \(n\) letters needed, then;

The # of single letter codes possible would be \(n\) itself;
The # of pair of distinct letters codes possible would be \(C^2_n\) (in alphabetical order);

We want \(C^2_n+n\geq{12}\) --> \(\frac{n(n-1)}{2}+n\geq{12}\) --> \(n(n-1)+2n\geq{24}\) --> \(n(n+1)\geq{24}\) --> \(n_{min}=5\).

Answer: B.

Hope it's clear.


Every time I see your explanation, problem becomes so easy, but when I tried my own, I hardly get the correct. How to improve my understanding on combination and Probability ?

Kudos [?]: [0], given: 4

Expert Post
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 42263

Kudos [?]: 132792 [0], given: 12373

Re: A researcher plans to identify each participant in a certain [#permalink]

Show Tags

New post 18 May 2014, 01:07
Expert's post
1
This post was
BOOKMARKED
gauravsaxena21 wrote:
Bunuel wrote:
sarb wrote:
A researcher plans to identify each participant in a certain medical experiment with a code consisting of either a single letter or a pair of distinct letters written in alphabetical order. What is the least number of letters that can be used if there are 12 participants, and each participant is to receive a different code?

A. 4
B. 5
C. 6
D. 7
E. 8


Say there are minimum of \(n\) letters needed, then;

The # of single letter codes possible would be \(n\) itself;
The # of pair of distinct letters codes possible would be \(C^2_n\) (in alphabetical order);

We want \(C^2_n+n\geq{12}\) --> \(\frac{n(n-1)}{2}+n\geq{12}\) --> \(n(n-1)+2n\geq{24}\) --> \(n(n+1)\geq{24}\) --> \(n_{min}=5\).

Answer: B.

Hope it's clear.


Every time I see your explanation, problem becomes so easy, but when I tried my own, I hardly get the correct. How to improve my understanding on combination and Probability ?


By studying theory and practicing.

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


Theory on probability problems: math-probability-87244.html

All DS probability problems to practice: search.php?search_id=tag&tag_id=33
All PS probability problems to practice: search.php?search_id=tag&tag_id=54

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


Hope this helps.
_________________

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

Kudos [?]: 132792 [0], given: 12373

Intern
Intern
avatar
Joined: 10 Jul 2014
Posts: 5

Kudos [?]: 5 [0], given: 16

Re: A researcher plans to identify each participant in a certain [#permalink]

Show Tags

New post 02 Sep 2014, 18:53
Bunuel wrote:
sarb wrote:
A researcher plans to identify each participant in a certain medical experiment with a code consisting of either a single letter or a pair of distinct letters written in alphabetical order. What is the least number of letters that can be used if there are 12 participants, and each participant is to receive a different code?

A. 4
B. 5
C. 6
D. 7
E. 8


Say there are minimum of \(n\) letters needed, then;

The # of single letter codes possible would be \(n\) itself;
The # of pair of distinct letters codes possible would be \(C^2_n\) (in alphabetical order);

We want \(C^2_n+n\geq{12}\) --> \(\frac{n(n-1)}{2}+n\geq{12}\) --> \(n(n-1)+2n\geq{24}\) --> \(n(n+1)\geq{24}\) --> \(n_{min}=5\).

Answer: B.

Hope it's clear.


I don't get how this is B instead of D. Aside from the formula, it says in alphabetical order, so how can you count AC, etc.?

Kudos [?]: 5 [0], given: 16

Expert Post
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 42263

Kudos [?]: 132792 [0], given: 12373

Re: A researcher plans to identify each participant in a certain [#permalink]

Show Tags

New post 02 Sep 2014, 20:16
Chin926926 wrote:
Bunuel wrote:
sarb wrote:
A researcher plans to identify each participant in a certain medical experiment with a code consisting of either a single letter or a pair of distinct letters written in alphabetical order. What is the least number of letters that can be used if there are 12 participants, and each participant is to receive a different code?

A. 4
B. 5
C. 6
D. 7
E. 8


Say there are minimum of \(n\) letters needed, then;

The # of single letter codes possible would be \(n\) itself;
The # of pair of distinct letters codes possible would be \(C^2_n\) (in alphabetical order);

We want \(C^2_n+n\geq{12}\) --> \(\frac{n(n-1)}{2}+n\geq{12}\) --> \(n(n-1)+2n\geq{24}\) --> \(n(n+1)\geq{24}\) --> \(n_{min}=5\).

Answer: B.

Hope it's clear.


I don't get how this is B instead of D. Aside from the formula, it says in alphabetical order, so how can you count AC, etc.?


AC is in alphabetical order while CA is not.
_________________

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

Kudos [?]: 132792 [0], given: 12373

Intern
Intern
avatar
Joined: 30 Jul 2014
Posts: 2

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

Re: A researcher plans to identify each participant in a certain [#permalink]

Show Tags

New post 10 Sep 2014, 11:41
Hi Bunuel,

I'm confused on when you show that n! = (n-2)!*(n-1)*n...why is n! only limited to those 3 factors? I guess the question is why do you start at (n-2)!?

Thanks.!


\(C^2_n=\frac{n!}{2!(n-2)!}\). Now, notice that \(n!=(n-2)!*(n-1)*n\), hence \(C^2_n=\frac{n!}{2!(n-2)!}=\frac{(n-2)!*(n-1)*n}{2!(n-2)!}=\frac{(n-1)n}{2}\).

Hope it's clear.

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

Expert Post
1 KUDOS received
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 42263

Kudos [?]: 132792 [1], given: 12373

Re: A researcher plans to identify each participant in a certain [#permalink]

Show Tags

New post 10 Sep 2014, 11:54
1
This post received
KUDOS
Expert's post
1
This post was
BOOKMARKED
kevn1115 wrote:
Hi Bunuel,

I'm confused on when you show that n! = (n-2)!*(n-1)*n...why is n! only limited to those 3 factors? I guess the question is why do you start at (n-2)!?

Thanks.!


\(C^2_n=\frac{n!}{2!(n-2)!}\). Now, notice that \(n!=(n-2)!*(n-1)*n\), hence \(C^2_n=\frac{n!}{2!(n-2)!}=\frac{(n-2)!*(n-1)*n}{2!(n-2)!}=\frac{(n-1)n}{2}\).

Hope it's clear.


n! is the product of positive integers from 1 to n, inclusive: n! = 1*2*...*(n-4)*(n-3)(n-2)(n-1)n. To simplify \(\frac{n!}{2!(n-2)!}\) I wrote n! as (n-2)!*(n-1)*n this enables us to reduce by (n-2)! to get \(\frac{(n-1)n}{2}\).

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

Kudos [?]: 132792 [1], given: 12373

Intern
Intern
avatar
Joined: 30 Jul 2014
Posts: 2

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

Re: A researcher plans to identify each participant in a certain [#permalink]

Show Tags

New post 10 Sep 2014, 12:06
Thanks Bunuel. I'm sorry, but I'm still not following.

Why couldn't you simplify n! as (n-4)!*(n-3)!*(n-2)!*(n-1)!*n? What tells you that n! starts with (n-2)! and not a smaller number like (n-3)! or further?

What is the significance about starting with "2" using (n-2)! and working your way down to "n" when simplifying this?

Thanks.

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

Expert Post
1 KUDOS received
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 42263

Kudos [?]: 132792 [1], given: 12373

A researcher plans to identify each participant in a certain [#permalink]

Show Tags

New post 10 Sep 2014, 12:21
1
This post received
KUDOS
Expert's post
1
This post was
BOOKMARKED
kevn1115 wrote:
Thanks Bunuel. I'm sorry, but I'm still not following.

Why couldn't you simplify n! as (n-4)!*(n-3)!*(n-2)!*(n-1)!*n? What tells you that n! starts with (n-2)! and not a smaller number like (n-3)! or further?

What is the significance about starting with "2" using (n-2)! and working your way down to "n" when simplifying this?

Thanks.


I think you need to brush up fundamentals.

The factorial of a non-negative integer n, denoted by \(n!\), is the product of all positive integers less than or equal to n.

For example: \(4!=1*2*3*4=24\).

Now, consider this: \(10! = 1*2*3*...*8*9*10\) can also be written as \(9!*10 = (1*2*3*...*8*9)*10 = 10!\) or \(8!*9*10 = (1*2*3*...*8)*9*10 = 10!\), or \(7!*8*9*10 = (1*2*3*...*7)*8*9*10 = 10!\) ...

Similarly \(n!\) (n factorial) can be written as \((n-2)!*(n-1)*n\) (n-2 factorial multiplied by (n-1)*n). We need to write it that way in order to be able to reduce by \((n-2)!\).

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

Kudos [?]: 132792 [1], given: 12373

Intern
Intern
User avatar
Joined: 25 Jun 2014
Posts: 46

Kudos [?]: 23 [0], given: 187

Re: A researcher plans to identify each participant in a certain [#permalink]

Show Tags

New post 22 Nov 2014, 01:47
Hi Bunuel,
if the same question leaves out the requirement of arranging with alphabetical order, the answer will be
n+ n(n-1) > 12
n2>12 => n= 4
Is this correct?

Kudos [?]: 23 [0], given: 187

Expert Post
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 42263

Kudos [?]: 132792 [0], given: 12373

Re: A researcher plans to identify each participant in a certain [#permalink]

Show Tags

New post 22 Nov 2014, 06:44
vietnammba wrote:
Hi Bunuel,
if the same question leaves out the requirement of arranging with alphabetical order, the answer will be
n+ n(n-1) > 12
n2>12 => n= 4
Is this correct?


Yes.

Check other similar questions: Constructing Numbers, Codes and Passwords.
_________________

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

Kudos [?]: 132792 [0], given: 12373

Director
Director
User avatar
Joined: 10 Mar 2013
Posts: 591

Kudos [?]: 479 [0], given: 200

Location: Germany
Concentration: Finance, Entrepreneurship
GMAT 1: 580 Q46 V24
GPA: 3.88
WE: Information Technology (Consulting)
GMAT ToolKit User
A researcher plans to identify each participant in a certain [#permalink]

Show Tags

New post 14 Dec 2014, 06:58
My solution was based on answers:
(A) 4 Letters: 4 Letters + 4!/2!*2! = 10 --> So it is not enough (2! means when out of 4 letters there are 2 letters in the code and 2 letters aren't)
(B) 5 Letters: 5 Letters + 5!/3!*2! = 15 --> So it's enough to build a code for 12 Persons

Solving with factorials, I calculated the amount of combinations of 2 distinct Letters for 4 or 5 Letters, so by 5 Letters, if 2 letters are in the code 3 of them are not --> 2! ans 3!.

I hope my solution was clear enough for you
_________________

When you’re up, your friends know who you are. When you’re down, you know who your friends are.

Share some Kudos, if my posts help you. Thank you !

800Score ONLY QUANT CAT1 51, CAT2 50, CAT3 50
GMAT PREP 670
MGMAT CAT 630
KAPLAN CAT 660

Kudos [?]: 479 [0], given: 200

Senior Manager
Senior Manager
User avatar
Status: Math is psycho-logical
Joined: 07 Apr 2014
Posts: 437

Kudos [?]: 141 [0], given: 169

Location: Netherlands
GMAT Date: 02-11-2015
WE: Psychology and Counseling (Other)
Re: A researcher plans to identify each participant in a certain [#permalink]

Show Tags

New post 26 Dec 2014, 06:40
I did this:

I also started by thinking than 12 single letters give 12 diffferent codes.
So, to use less, 12/2= 6 letters, that combined can give 12 codes (or more).

Then, I just created a letter matrix like this, using only the 6 first letters of the alphabet, and started selecting:
A B C D E F --> I 1st created the first row (6 letters max) and recreated the sequence beneath each letter
A A A A A A
B B B B B B
C C C C C C
D D D D D D
E E E E E E
F F F F F F

Then, I started selecting, using the least number of letters possible:
I selected the 1st row A, then the B beneath it --> A, AB (I deleted the second single A, as it is not allowed; same for rest)
I selected the 1st row B, then C beneath it --> B, BC
I selected the 1st row C, then the A beneath it (not to use more letters) --> C, AC
I selected 1st row D, then A,B,C beneath it (again using as much as the previous) --> D, AD, BD,CD
I selected 1st row E, then A beneath it (for the last 2 digits) --> E, AE.

Now I have 12 numbers, consisting either of one or two digits, in alphabetical order. I used 5 letters.

It can happen quite quickly, because having the matrix down on paper you can just cross out or select letters.

Kudos [?]: 141 [0], given: 169

Expert Post
5 KUDOS received
EMPOWERgmat Instructor
User avatar
P
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 10128

Kudos [?]: 3517 [5], given: 173

Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: 340 Q170 V170
Re: A researcher plans to identify each participant in a certain [#permalink]

Show Tags

New post 27 Dec 2014, 11:22
5
This post received
KUDOS
Expert's post
1
This post was
BOOKMARKED
Hi All,

In these sorts of questions, when the answer choices are relatively small, it's often fairly easy to "brute force" the correct answer and avoid complicated calculations entirely.

BrainLab's idea to just "map out" the possibilities is a relatively simple, effective approach. Since we're asked for the LEAST number of letters that will give us 12 unique codes, we start with Answer A.

If we had 4 letters: A, B, C, D

1-letter codes: A, B, C, D
2-letter alphabetical codes: AB, AC, AD, BC, BD, CD
Total Codes = 4 + 6 = 10

This result is TOO LOW.

From here, you know that we just need 2 more codes, so adding 1 more letter would give us those extra codes (and more)...but here's the proof that it happens....

If we had 5 letters: A, B, C, D, E

1-letter codes: A, B, C, D, E
2-letter alphabetical codes: AB, AC, AD, AE, BC, BD, BE, CD, CE, DE
Total Codes = 5 + 10 = 15 codes

Final Answer:
[Reveal] Spoiler:
B


GMAT assassins aren't born, they're made,
Rich
_________________

760+: Learn What GMAT Assassins Do to Score at the Highest Levels
Contact Rich at: Rich.C@empowergmat.com

Rich Cohen

Co-Founder & GMAT Assassin

Special Offer: Save $75 + GMAT Club Tests Free
  Official GMAT Exam Packs + 70 Pt. Improvement Guarantee
www.empowergmat.com/

***********************Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!***********************

Kudos [?]: 3517 [5], given: 173

Intern
Intern
User avatar
Status: Targeting 750
Joined: 13 Aug 2014
Posts: 6

Kudos [?]: 5 [0], given: 13

Location: Croatia (Hrvatska)
WE: Corporate Finance (Energy and Utilities)
GMAT ToolKit User
A researcher plans to identify each participant in a certain [#permalink]

Show Tags

New post 08 Feb 2015, 09:44
Bunuel wrote:
kevn1115 wrote:
Hi Bunuel,

I'm confused on when you show that n! = (n-2)!*(n-1)*n...why is n! only limited to those 3 factors? I guess the question is why do you start at (n-2)!?

Thanks.!


\(C^2_n=\frac{n!}{2!(n-2)!}\). Now, notice that \(n!=(n-2)!*(n-1)*n\), hence \(C^2_n=\frac{n!}{2!(n-2)!}=\frac{(n-2)!*(n-1)*n}{2!(n-2)!}=\frac{(n-1)n}{2}\).

Hope it's clear.


n! is the product of positive integers from 1 to n, inclusive: n! = 1*2*...*(n-4)*(n-3)(n-2)(n-1)n. To simplify \(\frac{n!}{2!(n-2)!}\) I wrote n! as (n-2)!*(n-1)*n this enables us to reduce by (n-2)! to get \(\frac{(n-1)n}{2}\).

Hope it's clear.


Hi bunuel!

Sorry for opening the old topic but my question is why are you and everyone else writing the "c(n, k)" upside down. Have a look at the screenshot of the gmat club math book and the equation everyone is referring to.

Thanks!
Attachments

image.jpg
image.jpg [ 366.12 KiB | Viewed 1230 times ]

Kudos [?]: 5 [0], given: 13

A researcher plans to identify each participant in a certain   [#permalink] 08 Feb 2015, 09:44

Go to page   Previous    1   2   3   4    Next  [ 72 posts ] 

Display posts from previous: Sort by

A researcher plans to identify each participant in a certain

  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 | 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®.