It is currently 22 Jun 2017, 07:47

### 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

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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# A certain club has 20 members. What is the ratio of the memb

Author Message
TAGS:

### Hide Tags

Manager
Joined: 02 Sep 2008
Posts: 103
A certain club has 20 members. What is the ratio of the memb [#permalink]

### Show Tags

25 Mar 2009, 23:04
9
This post was
BOOKMARKED
00:00

Difficulty:

15% (low)

Question Stats:

73% (02:04) correct 27% (00:59) wrong based on 582 sessions

### HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

A certain club has 20 members. What is the ratio of the member of 5-member committees that can be formed from the members of the club to the number of 4-member committees that can be formed from the members of the club?

A. 16 to 1
B. 15 to 1
C. 16 to 5
D. 15 to 6
E. 5 to 4
[Reveal] Spoiler: OA
Director
Joined: 01 Apr 2008
Posts: 880
Name: Ronak Amin
Schools: IIM Lucknow (IPMX) - Class of 2014

### Show Tags

25 Mar 2009, 23:17
1
This post was
BOOKMARKED
C.
20C5/20C4 = 16/5
milind1979 wrote:
A certain club has 20 members. What is the ratio of the member of 5-member
committees that can be formed from the members of the club to the number of 4-member
committees that can be formed from the members of the club?
A. 16 to 1
B. 15 to 1
C. 16 to 5
D. 15 to 6
E. 5 to 4
Retired Moderator
Status: Darden Class of 2013
Joined: 28 Jul 2009
Posts: 1836
Schools: University of Virginia

### Show Tags

24 Apr 2010, 09:38
brinng back an old post.

I somehow can't get 20C5/20C4 to be 16/5; I keep on getting 8/5

Also, for this question, why can't we just do the following:
(20*19*18*17*16)/(20*19*18*17) = 16/1?
_________________
Manager
Joined: 27 Dec 2009
Posts: 167

### Show Tags

24 Apr 2010, 09:47
1
This post was
BOOKMARKED
bakfed wrote:
brinng back an old post.

I somehow can't get 20C5/20C4 to be 16/5; I keep on getting 8/5

Also, for this question, why can't we just do the following:
(20*19*18*17*16)/(20*19*18*17) = 16/1?

20C5 = FACT[20] / FACT[5]*FACT[15]
20C4 = FACT[20] / FACT[4]*FACT[16]

20C5 / 20C4 = 16/5
Math Expert
Joined: 02 Sep 2009
Posts: 39582

### Show Tags

24 Apr 2010, 10:05
4
KUDOS
Expert's post
1
This post was
BOOKMARKED
bakfed wrote:
brinng back an old post.

I somehow can't get 20C5/20C4 to be 16/5; I keep on getting 8/5

Also, for this question, why can't we just do the following:
(20*19*18*17*16)/(20*19*18*17) = 16/1?

$$\frac{C^5_{20}}{C^4_{20}}=\frac{20!}{15!*5!}*\frac{16!*4!}{20!}=\frac{16}{5}$$

Second question:

20*19*18*17*16 does not give you # of 5 member committees out of 20. You need to divide this by 5! to get rid of the repetitions (factorial correction). The same for 20*19*18*17, you should divide this by 4!.

Take another example: how many committees of 2 can be formed out of A, B and C?

AB
AC
BC
Only 3, which is $$C^2_3=3$$.

But the way you are doing you'd get 3*2=6. This number has repetitions so we should divide it by 2! --> 6/2!=3.

Hope it's clear.
_________________
Manager
Joined: 27 Dec 2009
Posts: 167

### Show Tags

24 Apr 2010, 10:11
bakfed wrote:
brinng back an old post.

Also, for this question, why can't we just do the following:
(20*19*18*17*16)/(20*19*18*17) = 16/1?

20*19*18*17*16 is giving all possible ways of selecting 5 people. It also includes the order of a particular selection into account. In this case it does not mater if you select a member first ,second or third as long as it is selected in the group. Therefore, the 20*19*18*17*16 number has repetition of the same group in 5 ways and you have to divide by FACT[5]. Similarly for (20*19*18*17) by FACT[4]
Manager
Joined: 13 Jul 2010
Posts: 167

### Show Tags

19 Dec 2010, 15:46
I initially got this wrong, but see that this is just 20!/15!5! *16!4!/20! the 20! cancels out and you have 16!4!/15!5! which leaves 16/5.
Manager
Joined: 14 Jun 2011
Posts: 85

### Show Tags

29 Aug 2013, 11:11
Bunuel wrote:
bakfed wrote:
brinng back an old post.

I somehow can't get 20C5/20C4 to be 16/5; I keep on getting 8/5

Also, for this question, why can't we just do the following:
(20*19*18*17*16)/(20*19*18*17) = 16/1?

$$\frac{C^5_{20}}{C^4_{20}}=\frac{20!}{15!*5!}*\frac{16!*4!}{20!}=\frac{16}{5}$$

Second question:

20*19*18*17*16 does not give you # of 5 member committees out of 20. You need to divide this by 5! to get rid of the repetitions (factorial correction). The same for 20*19*18*17, you should divide this by 4!.

Take another example: how many committees of 2 can be formed out of A, B and C?

AB
AC
BC
Only 3, which is $$C^2_3=3$$.

But the way you are doing you'd get 3*2=6. This number has repetitions so we should divide it by 2! --> 6/2!=3.

Hope it's clear.

when should we use 20*19*18*17*16 way and when the second way?
_________________

Kudos always encourages me

Math Expert
Joined: 02 Sep 2009
Posts: 39582

### Show Tags

31 Aug 2013, 06:38
Expert's post
1
This post was
BOOKMARKED
swati007 wrote:
Bunuel wrote:
bakfed wrote:
brinng back an old post.

I somehow can't get 20C5/20C4 to be 16/5; I keep on getting 8/5

Also, for this question, why can't we just do the following:
(20*19*18*17*16)/(20*19*18*17) = 16/1?

$$\frac{C^5_{20}}{C^4_{20}}=\frac{20!}{15!*5!}*\frac{16!*4!}{20!}=\frac{16}{5}$$

Second question:

20*19*18*17*16 does not give you # of 5 member committees out of 20. You need to divide this by 5! to get rid of the repetitions (factorial correction). The same for 20*19*18*17, you should divide this by 4!.

Take another example: how many committees of 2 can be formed out of A, B and C?

AB
AC
BC
Only 3, which is $$C^2_3=3$$.

But the way you are doing you'd get 3*2=6. This number has repetitions so we should divide it by 2! --> 6/2!=3.

Hope it's clear.

when should we use 20*19*18*17*16 way and when the second way?

The first case is applicable when the order matters, when, for example, we have member #1, member #2, ...
_________________
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 15911
Re: A certain club has 20 members. What is the ratio of the memb [#permalink]

### Show Tags

16 Feb 2015, 21:51
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.
_________________
VP
Joined: 09 Jun 2010
Posts: 1412
Re: A certain club has 20 members. What is the ratio of the memb [#permalink]

### Show Tags

03 May 2015, 21:29
dont remember the formular, gmat dose not test.

to take 5
20*19*18*17*16/5!

Princeton review book is very exccelent at explanation of combination and counting. no book can compete Princeton book in this section. read it to master combination and counting. this section is only a few pages and can be learn in a few hours.

this explanation in the book is concise, deep and comprehensive.
_________________

visit my facebook to help me.
on facebook, my name is: thang thang thang

Director
Status: Verbal Forum Moderator
Joined: 17 Apr 2013
Posts: 613
Location: India
GMAT 1: 710 Q50 V36
GMAT 2: 750 Q51 V41
GMAT 3: 790 Q51 V49
GPA: 3.3
Re: A certain club has 20 members. What is the ratio of the memb [#permalink]

### Show Tags

27 Jul 2015, 06:01
Isnt the Question doubtful?

They didnt mentioned that first we have to select 5 members then 4 from the remaining 15?
_________________

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

Math Forum Moderator
Joined: 20 Mar 2014
Posts: 2642
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
Re: A certain club has 20 members. What is the ratio of the memb [#permalink]

### Show Tags

27 Jul 2015, 06:08
honchos wrote:
Isnt the Question doubtful?

They didnt mentioned that first we have to select 5 members then 4 from the remaining 15?

That is exactly the reason why you need to treat "selecting 5 out of 20" and "selecting 4 out of 20" as 2 independent events. Per the question, there is no overlap in the 2 cases.

Thus the final answer = 20C5 / 20C4 = 16/5
_________________

Thursday with Ron updated list as of July 1st, 2015: http://gmatclub.com/forum/consolidated-thursday-with-ron-list-for-all-the-sections-201006.html#p1544515
Inequalities tips: http://gmatclub.com/forum/inequalities-tips-and-hints-175001.html
Debrief, 650 to 750: http://gmatclub.com/forum/650-to-750-a-10-month-journey-to-the-score-203190.html

SVP
Joined: 08 Jul 2010
Posts: 1731
Location: India
GMAT: INSIGHT
WE: Education (Education)
Re: A certain club has 20 members. What is the ratio of the memb [#permalink]

### Show Tags

27 Jul 2015, 06:13
honchos wrote:
Isnt the Question doubtful?

They didnt mentioned that first we have to select 5 members then 4 from the remaining 15?

The question is absolutely clear and holds no ambiguity in the language

Question : A certain club has 20 members. What is the ratio of the member of 5-member committees that can be formed from the members of the club to the number of 4-member committees that can be formed from the members of the club?"

The two actions (making the committee of 5 members AND making the committee of 4 embers) are to be treated independently.

No. of Ways to make 5 member committee out of 20 members = 20C5 = 15504

No. of Ways to make 4 member committee out of 20 members = 20C4 = 4845

Required ratio = 15504 / 4845 = 16/5
_________________

Prosper!!!
GMATinsight
Bhoopendra Singh and Dr.Sushma Jha
e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772
Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi
http://www.GMATinsight.com/testimonials.html

22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION

Director
Status: Verbal Forum Moderator
Joined: 17 Apr 2013
Posts: 613
Location: India
GMAT 1: 710 Q50 V36
GMAT 2: 750 Q51 V41
GMAT 3: 790 Q51 V49
GPA: 3.3
Re: A certain club has 20 members. What is the ratio of the memb [#permalink]

### Show Tags

27 Jul 2015, 06:22
Engr2012 wrote:
honchos wrote:
Isnt the Question doubtful?

They didnt mentioned that first we have to select 5 members then 4 from the remaining 15?

That is exactly the reason why you need to treat "selecting 5 out of 20" and "selecting 4 out of 20" as 2 independent events. Per the question, there is no overlap in the 2 cases.

Thus the final answer = 20C5 / 20C4 = 16/5

Thanks, My doubt is still the same. They didn't mentioned that two events are mutually independent or dependent.
This is a very easy question, but I was spinned that how should I interpret the language - Mutually independent or independent.
_________________

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

Math Forum Moderator
Joined: 20 Mar 2014
Posts: 2642
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
A certain club has 20 members. What is the ratio of the memb [#permalink]

### Show Tags

27 Jul 2015, 06:25
Dependent events is a very big information and will thus always be mentioned if that is the case. If not given, you have to treat the 2 events independently. You are over analyzing the question. The language is perfectly fine.
_________________

Thursday with Ron updated list as of July 1st, 2015: http://gmatclub.com/forum/consolidated-thursday-with-ron-list-for-all-the-sections-201006.html#p1544515
Inequalities tips: http://gmatclub.com/forum/inequalities-tips-and-hints-175001.html
Debrief, 650 to 750: http://gmatclub.com/forum/650-to-750-a-10-month-journey-to-the-score-203190.html

SVP
Joined: 08 Jul 2010
Posts: 1731
Location: India
GMAT: INSIGHT
WE: Education (Education)
Re: A certain club has 20 members. What is the ratio of the memb [#permalink]

### Show Tags

27 Jul 2015, 06:29
honchos wrote:
Engr2012 wrote:
honchos wrote:
Isnt the Question doubtful?

They didnt mentioned that first we have to select 5 members then 4 from the remaining 15?

That is exactly the reason why you need to treat "selecting 5 out of 20" and "selecting 4 out of 20" as 2 independent events. Per the question, there is no overlap in the 2 cases.

Thus the final answer = 20C5 / 20C4 = 16/5

Thanks, My doubt is still the same. They didn't mentioned that two events are mutually independent or dependent.
This is a very easy question, but I was spinned that how should I interpret the language - Mutually independent or independent.

The question did NOT mentioned that "Team of 4 has to be selected out of the REMAINING"

If the word "remaining" has not been used then it is Independent event case.

I hope it helps1
_________________

Prosper!!!
GMATinsight
Bhoopendra Singh and Dr.Sushma Jha
e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772
Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi
http://www.GMATinsight.com/testimonials.html

22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION

Manager
Joined: 18 Feb 2015
Posts: 93
Re: A certain club has 20 members. What is the ratio of the memb [#permalink]

### Show Tags

09 Feb 2016, 14:33
Hello guys,

Can I please ask how 20C5 / 20C4 has been solved to get the fraction 16/5?
I know that one way to solve this is by doing the long multiplication thing but Brunnel has calculated it in a simplified manner. Can someone please write down the steps?

Thanks,
SVP
Joined: 08 Jul 2010
Posts: 1731
Location: India
GMAT: INSIGHT
WE: Education (Education)
Re: A certain club has 20 members. What is the ratio of the memb [#permalink]

### Show Tags

10 Feb 2016, 01:41
1
KUDOS
Expert's post
HarveyKlaus wrote:
Hello guys,

Can I please ask how 20C5 / 20C4 has been solved to get the fraction 16/5?
I know that one way to solve this is by doing the long multiplication thing but Brunnel has calculated it in a simplified manner. Can someone please write down the steps?

Thanks,

20C5 = 20!/[5! * 15!] = 20*19*18*17*16/5! = 19*3*17*16

20C4 = 20!/[4! * 16!] = 20*19*18*17/4! = 5*19*3*17

20C5 / 20C4 = 19*3*17*16 / 5*19*3*17 = 16/5

I hope this helps!!!
_________________

Prosper!!!
GMATinsight
Bhoopendra Singh and Dr.Sushma Jha
e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772
Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi
http://www.GMATinsight.com/testimonials.html

22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION

Optimus Prep Instructor
Joined: 06 Nov 2014
Posts: 1831
Re: A certain club has 20 members. What is the ratio of the memb [#permalink]

### Show Tags

06 Jun 2016, 08:41
5 member committee = 20C5
4 member committee = 20C4

Ratio = 20C5 / 20C4 = (20!/15!*5!) * (16!*4!/20!) = 16: 5

Correct option: C
_________________

# Janielle Williams

Customer Support

Special Offer: $80-100/hr. Online Private Tutoring GMAT On Demand Course$299
Free Online Trial Hour

Re: A certain club has 20 members. What is the ratio of the memb   [#permalink] 06 Jun 2016, 08:41

Go to page    1   2    Next  [ 22 posts ]

Similar topics Replies Last post
Similar
Topics:
8 A local club has between 24 and 57 members. The members of the club ca 9 10 May 2017, 12:39
30 A certain club has exactly 5 new members at the end of its 12 13 Sep 2016, 11:51
4 A certain club has 10 members, including Harry. One of the 8 25 Jul 2012, 21:13
145 A certain club has 10 members, including Harry. One of the 33 31 May 2017, 22:01
1 A certain team has 12 members, including Joey. A three-membe 13 14 Dec 2011, 10:58
Display posts from previous: Sort by