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

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

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New post 03 Jul 2013, 06:27
10
00:00
A
B
C
D
E

Difficulty:

  85% (hard)

Question Stats:

39% (02:24) correct 61% (01:30) wrong based on 28 sessions

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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?

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

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New post 24 Mar 2014, 02:12
2
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?

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

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New post 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?

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

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New post 28 Apr 2014, 10:11
3
1
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?

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

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New post 20 Mar 2019, 15:32
Hi guys ,I dont' understand why you considered them different objects
I think the votes and the attractions should be considered identical ,
let's count the total way of placing 6 votes into 4 rooms with no constraints , so any room can get a zero or any number ,
(we don't care who voted for which , we car only about the number of votes each room gets )

then let's find how many ways all rooms get at least one vote


not sure if my approach is correct , please correct me :

6 identical votes need to go into 4 identical rooms , first let's find the total number with no constraints :
let's make 3 walls , vvvvvv||| ,9 objects , let's choose 3 of them to place our walls : 9C3=84 ways

now , let's give each room a vote at first
we are left with 2 votes and 4 rooms , let's make 3 walls , vv||| , ok : 5 objects , let's choose 3 of them to place our walls :
5C3 = 10 ways
so p = 10/84
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Re: Six friends live in the city of Monrovia. There are four nat  [#permalink]

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New post 23 Mar 2019, 13:00
Bunuel

I am not able to understand here that why does it matter who has voted for which place. why are we multiplying by 6C4 here?? the question simply asks that each place should have atleast one vote. why does it matter from whom that vote has been casted???
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Re: Six friends live in the city of Monrovia. There are four nat   [#permalink] 23 Mar 2019, 13:00
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