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 500,000 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:

Six friends live in the city of Monrovia. There are four nat [#permalink]

Show Tags

03 Jul 2013, 05:27

10

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

85% (hard)

Question Stats:

48% (03:12) correct
53% (02:12) wrong based on 40 sessions

HideShow timer Statistics

Found this question in one of the Veritas Prep blog posts...

Six friends live in the city of Monrovia. There are four natural attractions around Monrovia – a waterfall, a safari, a lake and some caves. The friends decide to take a vacation together at one of these attractions. To select the attraction, each one of them votes for one of the attractions. What is the probability that each attraction gets at least one vote?

The answer is 1560/(4^6). I was wondering if we can get to this answer by removing the cases where [an "attraction" gets zero votes] can be subtracted from the [total number of cases].

Found this question in one of the Veritas Prep blog posts...

Six friends live in the city of Monrovia. There are four natural attractions around Monrovia – a waterfall, a safari, a lake and some caves. The friends decide to take a vacation together at one of these attractions. To select the attraction, each one of them votes for one of the attractions. What is the probability that each attraction gets at least one vote?

The answer is 1560/(4^6). I was wondering if we can get to this answer by removing the cases where [an "attraction" gets zero votes] can be subtracted from the [total number of cases].

Counting those is harder than a direct approach when we need to count only: {3, 1, 1, 1} --> \(\frac{4!}{3!}*C^3_6*3!=480\). {2, 2, 1, 1} --> \(\frac{4!}{2!2!}*C^2_6*C^2_4*C^1_2=1,080\).

The total number of ways in which the 6 friends can vote = 4*4*4*4*4*4 = 4^6.

Re: Six friends live in the city of Monrovia. There are four nat [#permalink]

Show Tags

28 Apr 2014, 04:35

Bunuel wrote:

v1gnesh wrote:

Found this question in one of the Veritas Prep blog posts...

Six friends live in the city of Monrovia. There are four natural attractions around Monrovia – a waterfall, a safari, a lake and some caves. The friends decide to take a vacation together at one of these attractions. To select the attraction, each one of them votes for one of the attractions. What is the probability that each attraction gets at least one vote?

The answer is 1560/(4^6). I was wondering if we can get to this answer by removing the cases where [an "attraction" gets zero votes] can be subtracted from the [total number of cases].

Counting those is harder than a direct approach when we need to count only: {3, 1, 1, 1} --> \(\frac{4!}{3!}*C^3_6*3!=480\). {2, 2, 1, 1} --> \(\frac{4!}{2!2!}*C^2_6*C^2_4*C^1_2=1,080\).

The total number of ways in which the 6 friends can vote = 4*4*4*4*4*4 = 4^6.

P = (480 + 1,080)/4^6 = 1560/4^6.

Hope it helps.

Bunuel

Please Elaborate how you got the 3 Equations :

Counting those is harder than a direct approach when we need to count only: {3, 1, 1, 1} --> \(\frac{4!}{3!}*C^3_6*3!=480\). {2, 2, 1, 1} --> \(\frac{4!}{2!2!}*C^2_6*C^2_4*C^1_2=1,080\).

The total number of ways in which the 6 friends can vote = 4*4*4*4*4*4 = 4^6.

Found this question in one of the Veritas Prep blog posts...

Six friends live in the city of Monrovia. There are four natural attractions around Monrovia – a waterfall, a safari, a lake and some caves. The friends decide to take a vacation together at one of these attractions. To select the attraction, each one of them votes for one of the attractions. What is the probability that each attraction gets at least one vote?

The answer is 1560/(4^6). I was wondering if we can get to this answer by removing the cases where [an "attraction" gets zero votes] can be subtracted from the [total number of cases].

Counting those is harder than a direct approach when we need to count only: {3, 1, 1, 1} --> \(\frac{4!}{3!}*C^3_6*3!=480\). {2, 2, 1, 1} --> \(\frac{4!}{2!2!}*C^2_6*C^2_4*C^1_2=1,080\).

The total number of ways in which the 6 friends can vote = 4*4*4*4*4*4 = 4^6.

P = (480 + 1,080)/4^6 = 1560/4^6.

Hope it helps.

Bunuel

Please Elaborate how you got the 3 Equations :

Counting those is harder than a direct approach when we need to count only: {3, 1, 1, 1} --> \(\frac{4!}{3!}*C^3_6*3!=480\). {2, 2, 1, 1} --> \(\frac{4!}{2!2!}*C^2_6*C^2_4*C^1_2=1,080\).

The total number of ways in which the 6 friends can vote = 4*4*4*4*4*4 = 4^6.

Thankyou!

For each attraction to get at least one vote we should have the following distributions of 6 votes: {3, 1, 1, 1} or {2, 2, 1, 1}.

A - B - C - D (attractions) {3, 1, 1, 1} --> \(\frac{4!}{3!}*C^3_6*3!=480\), where: \(\frac{4!}{3!}\) is the # assignments of {3, 1, 1, 1} to attractions (A gets 3, B gets 1, C gets 1, D gets 1; A gets 1, B gets 3, C gets 1, D gets 1; A gets 1, B gets 1, C gets 3, D gets 1; A gets 1, B gets 1, C gets 1, D gets 3); \(C^3_6\) is selecting the 3 people who will vote for the attraction with 3 votes; 3! is the distribution of other 3 votes among the remaining attractions.

The same logic applies to {2, 2, 1, 1} case.

As for 4^6: each friend can vote in 4 ways (for waterfall, safari, lake or caves). So, the total number of ways in which the 6 friends can vote = 4*4*4*4*4*4 = 4^6.

Re: Six friends live in the city of Monrovia. There are four nat [#permalink]

Show Tags

03 Jan 2016, 03:39

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.
_________________

Campus visits play a crucial role in the MBA application process. It’s one thing to be passionate about one school but another to actually visit the campus, talk...

Its been long time coming. I have always been passionate about poetry. It’s my way of expressing my feelings and emotions. And i feel a person can convey...

Marty Cagan is founding partner of the Silicon Valley Product Group, a consulting firm that helps companies with their product strategy. Prior to that he held product roles at...

Written by Scottish historian Niall Ferguson , the book is subtitled “A Financial History of the World”. There is also a long documentary of the same name that the...