Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 27 Oct 2016, 22:07

### 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

# Events & Promotions

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

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

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

### Hide Tags

Intern
Joined: 26 Mar 2013
Posts: 12
Location: United States
Followers: 0

Kudos [?]: 12 [0], given: 3

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

### Show Tags

03 Jul 2013, 06:27
10
This post was
BOOKMARKED
00:00

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?

[Reveal] Spoiler: OE
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].
[Reveal] Spoiler: OA
 Veritas Prep GMAT Discount Codes Manhattan GMAT Discount Codes e-GMAT Discount Codes
Math Expert
Joined: 02 Sep 2009
Posts: 35327
Followers: 6649

Kudos [?]: 85885 [2] , given: 10260

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

### Show Tags

24 Mar 2014, 02:12
2
This post received
KUDOS
Expert's post
2
This post was
BOOKMARKED
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?

[Reveal] Spoiler: OE
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].

The cases when attractions get zero votes are:

{6, 0, 0, 0}
{5, 1, 0, 0}
{4, 2, 0, 0}
{4, 1, 1, 0}
{3, 2, 1, 0}
{3, 1, 1, 1}
{2, 2, 2, 0}
{2, 2, 1, 1}

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.
_________________
Intern
Joined: 22 Jun 2013
Posts: 45
Followers: 0

Kudos [?]: 40 [0], given: 132

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

### Show Tags

28 Apr 2014, 05:35
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?

[Reveal] Spoiler: OE
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].

The cases when attractions get zero votes are:

{6, 0, 0, 0}
{5, 1, 0, 0}
{4, 2, 0, 0}
{4, 1, 1, 0}
{3, 2, 1, 0}
{3, 1, 1, 1}
{2, 2, 2, 0}
{2, 2, 1, 1}

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!
Math Expert
Joined: 02 Sep 2009
Posts: 35327
Followers: 6649

Kudos [?]: 85885 [3] , given: 10260

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

### Show Tags

28 Apr 2014, 10:11
3
This post received
KUDOS
Expert's post
niyantg wrote:
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?

[Reveal] Spoiler: OE
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].

The cases when attractions get zero votes are:

{6, 0, 0, 0}
{5, 1, 0, 0}
{4, 2, 0, 0}
{4, 1, 1, 0}
{3, 2, 1, 0}
{3, 1, 1, 1}
{2, 2, 2, 0}
{2, 2, 1, 1}

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.

Hope it's clear.
_________________
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 12267
Followers: 542

Kudos [?]: 151 [0], given: 0

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

### Show Tags

03 Jan 2016, 04:39
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.
_________________
Re: Six friends live in the city of Monrovia. There are four nat   [#permalink] 03 Jan 2016, 04:39
Similar topics Replies Last post
Similar
Topics:
Four representatives are to be randomly chosen from a committee of six 2 28 Sep 2016, 03:00
1 Four friends, Saul, Peter, Quirinal, and Roderick, are pooling their 4 11 May 2016, 00:09
3 Will decides to attend a basketball game with four friends. 3 17 Jan 2014, 07:21
5 A group of six friends sit together and form a circle. They 3 14 Oct 2013, 07:05
Four female friends & four male friends will be pictured in 2 05 Apr 2012, 01:30
Display posts from previous: Sort by

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

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group and phpBB SEO 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®.