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.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

There are 5 chess amateurs playing in Villa's chess club [#permalink]
07 Feb 2012, 20:36

3

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

25% (medium)

Question Stats:

60% (01:37) correct
40% (00:37) wrong based on 156 sessions

There are 5 chess amateurs playing in Villa's chess club tournament. If each chess amateur plays with exactly 4 other amateurs, what is the total number of chess games possible to be played in the tournament?

There are 5 chess amateurs playing in Villa's chess club tournament. If each chess amateur plays with exactly 4 other amateurs, what is the total number of chess games possible to be played in the tournament?

A 10 B 20 C 40 D 60 E 120

There are different ways of approaching this question.

Method 1: Take the first amateur. He plays a game with each of the other four i.e. 4 games. Now take the second one. He has already played a game with the first one. He plays 3 games with the rest of the 3 amateurs i.e. 3 more games are played. Now take the third amateur. He has already played a game each with the first and the second amateur. Now he plays 2 games with the remaining 2 amateurs so 2 more games are played. Now go on to the fourth amateur. He has already played 3 games with the first 3 amateurs. He just needs to play a game with the last one i.e. 1 more game is played. The last amateur has already played 4 games. Total no of games = 4+3+2+1 = 10

Method 2: Each person is one participant of 4 games. So there are in all 4*5 = 20 instances of one participant games. But each game has 2 participants so total number of games = 20/2 = 10 _________________

There are 5 chess amateurs playing in Villa's chess club tournament. If each chess amateur plays with exactly 4 other amateurs, what is the total number of chess games possible to be played in the tournament?

A 10 B 20 C 40 D 60 E 120

The easiest way probably would be to realize that total # of games possible will be total # of different pairs possible out of 5 players (one game per one pair): \(C^2_5=\frac{5!}{3!*2!}=10\).

Re: There are 5 chess amateurs playing in Villa's chess club [#permalink]
08 Feb 2012, 05:16

1

This post received KUDOS

First player = 4 different games possible Second player = 3 different games possible (1 has already counted) third player = 2 different games possible fourht player = 1 different game possible

Re: There are 5 chess amateurs playing in Villa's chess club [#permalink]
07 Sep 2013, 19:54

Quote:

The easiest way probably would be to realize that total # of games possible will be total # of different pairs possible out of 5 players (one game per one pair): C^2_5=\frac{5!}{3!*2!}.

I'm a little confused with this method.

If I use the Combination formula - shouldn't it be (5!/(3!*2!) x 4(Since it's being played with 4 other people?) If not - what's the point of having that text in the question? Would appreciate any clarification!

Re: There are 5 chess amateurs playing in Villa's chess club [#permalink]
07 Sep 2013, 20:37

russ9 wrote:

Quote:

The easiest way probably would be to realize that total # of games possible will be total # of different pairs possible out of 5 players (one game per one pair): C^2_5=\frac{5!}{3!*2!}.

I'm a little confused with this method.

If I use the Combination formula - shouldn't it be (5!/(3!*2!) x 4(Since it's being played with 4 other people?) If not - what's the point of having that text in the question? Would appreciate any clarification!

Total number of ways of pairing people out of a total of 5 which is 5C2 implies that each person is paired with or plays exactly with the remaining 4 persons . So it is enough if you use the formula 5C2. _________________

Re: There are 5 chess amateurs playing in Villa's chess club [#permalink]
08 Sep 2013, 07:00

Expert's post

russ9 wrote:

Quote:

The easiest way probably would be to realize that total # of games possible will be total # of different pairs possible out of 5 players (one game per one pair): C^2_5=\frac{5!}{3!*2!}.

I'm a little confused with this method.

If I use the Combination formula - shouldn't it be (5!/(3!*2!) x 4(Since it's being played with 4 other people?) If not - what's the point of having that text in the question? Would appreciate any clarification!

The formula nCr gives you the number of distinct ways in which you can select r people out of n.

e.g. If there are 5 people and you need to select 2, you can do it in many ways.

e.g. A, B, C, D, E

A, B A, C A, D A, E B, C . . . etc

In how many ways can you select 2 people? In 5C2 ways = 5!/3!*2! ways. This gives us the number of games possible. So there is no reason to multiply it by 4. _________________

Re: There are 5 chess amateurs playing in Villa's chess club [#permalink]
11 Sep 2013, 20:45

Thanks Karishma and Sravna. I guess I was a little thrown off because I've seen this formula with just the numerator(using arbitrary numbers) - just 5! and not divided by anything, therefore the confusion as to when to divide by the remainder factorials vs. when to leave it alone. I'm assuming I divide only when there is a set number of ways that a pair can be formed?

On the flipside - if the question had broken this up into something similar to follows - "There were 5 people, 3 boys and 2 girls, and boys can only play against girls and vice versa. How many combinations are there?" How would that change the formula?

Re: There are 5 chess amateurs playing in Villa's chess club [#permalink]
12 Sep 2013, 01:33

russ9 wrote:

Thanks Karishma and Sravna. I guess I was a little thrown off because I've seen this formula with just the numerator(using arbitrary numbers) - just 5! and not divided by anything, therefore the confusion as to when to divide by the remainder factorials vs. when to leave it alone. I'm assuming I divide only when there is a set number of ways that a pair can be formed?

On the flipside - if the question had broken this up into something similar to follows - "There were 5 people, 3 boys and 2 girls, and boys can only play against girls and vice versa. How many combinations are there?" How would that change the formula?

Would it be 3C2 + 2C2? Thanks!

In the original problem, out of the 5 players any two can play irrespective of whether both are boys or both are girls. So you are selecting 2 out of 5. So it is 5C2.

In your problem, a boy can play only against a girl. So you need to first select a boy out of the 3 boys and a girl out of the 2 girls. The former can be done in 3C1 ways and the latter in 2C1 ways. The total number of ways is 3C1 * 2C1 = 6 ways. You need to multiply because each boy selected, can play against 2 girls. _________________

Re: There are 5 chess amateurs playing in Villa's chess club [#permalink]
12 Sep 2013, 02:40

Expert's post

1

This post was BOOKMARKED

russ9 wrote:

Thanks Karishma and Sravna. I guess I was a little thrown off because I've seen this formula with just the numerator(using arbitrary numbers) - just 5! and not divided by anything, therefore the confusion as to when to divide by the remainder factorials vs. when to leave it alone. I'm assuming I divide only when there is a set number of ways that a pair can be formed?

On the flipside - if the question had broken this up into something similar to follows - "There were 5 people, 3 boys and 2 girls, and boys can only play against girls and vice versa. How many combinations are there?" How would that change the formula?

Would it be 3C2 + 2C2? Thanks!

Do you mean 'boys play against boys and girls play against girls' or do you mean 'every game has one boy and one girl'?

The answer you have given assumes the first case but the question you have framed assumes the second!

If every game has one boy and one girl, you select a boy out of 3 in 3C1 ways and a girl out of 2 girls in 2C1 ways. You need to select both so you multiply 3*2 = 6 ways

If boys play against boys and girls against girls, you can select 2 boys out of 3 for a game in 3C2 ways or you can select 2 girls out of 2 in 2C2 ways. Total = 3C2 + 2C2 = 4 ways _________________

Re: There are 5 chess amateurs playing in Villa's chess club [#permalink]
21 Nov 2014, 23:35

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. _________________

Hey, everyone. After a hectic orientation and a weeklong course, Managing Groups and Teams, I have finally settled into the core curriculum for Fall 1, and have thus found...

MBA Acceptance Rate by Country Most top American business schools brag about how internationally diverse they are. Although American business schools try to make sure they have students from...

After I was accepted to Oxford I had an amazing opportunity to visit and meet a few fellow admitted students. We sat through a mock lecture, toured the business...