NEW HERE? LEARN MORE
closeclose
Last visit was: 26 Apr 2024, 02:56 It is currently 26 Apr 2024, 02:56
Decision Tracker
My Rewards
New posts
New comers' posts

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.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel

Two members of a club are to be selected to represent the club at a na

SORT BY:
Date
Kudos
Tags:
Find Similar Topics 
Show Tags
Hide Tags
avatar
arnijon90
Intern
Intern
Joined: 18 Mar 2012
Posts: 11
Own Kudos [?]: 24 [19]
Given Kudos: 3
Send PM
Two members of a club are to be selected to represent the club at a na [#permalink] New post   Updated on: 20 Sep 2019, 03:41
3
Kudos
16
Bookmarks
00:00
A
B
C
D
E

Difficulty:

5% (low)

Question Stats:

86% (01:35) correct 14% (01:28) wrong based on 414 sessions
Hide Show timer Statistics
Two members of a club are to be selected to represent the club at a national meeting. if there are 190 different possible selections of the 2 members, how many members does the club have?

A. 20
B. 27
C. 40
D. 57
E. 95

Show Spoiler
I know I can just plug in answer choices in to the formula n!/(n-r)!*r!=190

n=20


but I saw a explanation on this problem where they showed this step

n!/(n-2)! = n(n-1)

How is this step possible ?
ShowHide Answer
Official Answer
A

Originally posted by arnijon90 on 10 May 2012, 08:15.
Last edited by Bunuel on 20 Sep 2019, 03:41, edited 2 times in total.
Edited the question
Most Helpful Reply
User avatar
P
energetics
Senior Manager
Senior Manager
Joined: 05 Feb 2018
Posts: 312
Own Kudos [?]: 794 [5]
Given Kudos: 325
Send PM
Re: Two members of a club are to be selected to represent the club at a na [#permalink] New post   13 Feb 2019, 15:18
5
Kudos
nC2, or n(n-1)/2 = 190
n(n-1) = 380
We can use prime factorization: 380 = 2*2*5*19
Becomes pretty obvious that 19 is one of the numbers (n-1) and the other is n=20.
User avatar
L
Bunuel
Math Expert
Joined: 02 Sep 2009
Posts: 92929
Own Kudos [?]: 619100 [5]
Given Kudos: 81609
Send PM
CAT Tests Forum Quiz Mode
Re: Two members of a club are to be selected to represent the club at a na [#permalink] New post   10 May 2012, 08:28
1
Kudos
4
Bookmarks
Expert Reply
arnijon90 wrote:
Two members of a club are to be selected to represent the club at a national meeting. if there are 190 different possible selections of the 2 members, how many members does the club have?

A. 20
B. 27
C. 40
D. 57
E. 95

I know I can just plug in answer choices in to the formula n!/(n-r)!*r!=190

n=20


but I saw a explanation on this problem where they showed this step

n!/(n-2)! = n(n-1)

How is this step possible ?


Welcome to GMAT Club. Below is an answer to your question.

\(C^2_n=190\) --> \(\frac{n!}{2!*(n-2)!}=190\). Now, notice that \(n!=(n-2)!*(n-1)*n\), so \(\frac{(n-2)!*(n-1)*n}{2!*(n-2)!}=190\) --> \(\frac{(n-1)*n}{2}=190\) --> \((n-1)n=380\) --> \(n=20\).

Answer: A.

P.S. Please always post answer choices for PS questions. On the PS section one should always look at the answer choices before starts to solve a problem. They might often give a clue on how to approach the question.
General Discussion
avatar
nishantmehra01
Intern
Intern
Joined: 26 Nov 2011
Posts: 9
Own Kudos [?]: 4 [0]
Given Kudos: 0
Send PM
Re: Two members of a club are to be selected to represent the club at a na [#permalink] New post   21 Jun 2012, 01:30
I have done this problem by substituting numbers


We have n!/2!(n-2)! = 190

or n(n-1)/2 = 190


n=20 satisfies the equation


arnijon90 wrote:
Two members of a club are to be selected to represent the club at a national meeting. if there are 190 different possible selections of the 2 members, how many members does the club have?

A. 20
B. 27
C. 40
D. 57
E. 95

I know I can just plug in answer choices in to the formula n!/(n-r)!*r!=190

n=20


but I saw a explanation on this problem where they showed this step

n!/(n-2)! = n(n-1)

How is this step possible ?
User avatar
L
ScottTargetTestPrep
Target Test Prep Representative
Joined: 14 Oct 2015
Status:Founder & CEO
Affiliations: Target Test Prep
Posts: 18761
Own Kudos [?]: 22055 [2]
Given Kudos: 283
Location: United States (CA)
Send PM
Re: Two members of a club are to be selected to represent the club at a na [#permalink] New post   15 Feb 2019, 18:32
1
Kudos
1
Bookmarks
Expert Reply
arnijon90 wrote:
Two members of a club are to be selected to represent the club at a national meeting. if there are 190 different possible selections of the 2 members, how many members does the club have?

A. 20
B. 27
C. 40
D. 57
E. 95


The order of selecting the two representatives is of no importance; thus, we use combinations. Letting n = the total number of members of the club, we can create the equation:

nC2 = 190

nC2 = n!/[(n - 2)! x 2!] =[ n x (n-1) x (n-2) x (n-1) x … x 1] / {[(n-2) x (n-1) x … x 1] x 2!}

We see that all the factors in the numerator cancel with those in the denominator, except n x (n-1). Thus, we have:

(n)(n-1)/2 = 190

n^2 - n = 380

n^2 - n - 380 = 0

(n - 20)(n + 19) = 0

n = 20 or n = -19

Since n can’t be negative, then n must be 20.

Answer: A
User avatar
L
IanStewart
GMAT Tutor
Joined: 24 Jun 2008
Posts: 4128
Own Kudos [?]: 9245 [1]
Given Kudos: 91
 Q51  V47
Send PM
Re: Two members of a club are to be selected to represent the club at a na [#permalink] New post   16 Feb 2019, 16:35
1
Bookmarks
Expert Reply
(n)(n-1)/2! = 190
(n)(n-1) = 380
n^2 - n = 380

Since n is a fairly large positive integer, n^2 is much larger than n, so we want a value of n^2 that is very close to 380. Glancing at the answer choices, n = 20 is the only realistic candidate, and it's easy to plug it back in to confirm it works.
User avatar
bumpbot
Non-Human User
Joined: 09 Sep 2013
Posts: 32678
Own Kudos [?]: 822 [0]
Given Kudos: 0
Send PM
Re: Two members of a club are to be selected to represent the club at a na [#permalink] New post   15 Aug 2023, 14:09
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 Club Bot
gmatclubot
Re: Two members of a club are to be selected to represent the club at a na [#permalink]
15 Aug 2023, 14:09
Moderators:
Bunuel
Math Expert
92929 posts
yashikaaggarwal
Senior Moderator - Masters Forum
3137 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne
Main navigation
GMAT Resources
Partners
MBA Resources
www.gmac.com|www.mba.com | | GMAT Club Rules | Terms and Conditions