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

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Intern
Joined: 26 Mar 2013
Posts: 11
Location: United States
Six friends live in the city of Monrovia. There are four nat  [#permalink]

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03 Jul 2013, 06:27
23
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Difficulty:

95% (hard)

Question Stats:

38% (02:30) correct 62% (01:45) wrong based on 39 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].
Math Expert
Joined: 02 Sep 2009
Posts: 59674
Re: Six friends live in the city of Monrovia. There are four nat  [#permalink]

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24 Mar 2014, 02:12
4
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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Posts: 35
Re: Six friends live in the city of Monrovia. There are four nat  [#permalink]

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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!
Math Expert
Joined: 02 Sep 2009
Posts: 59674
Re: Six friends live in the city of Monrovia. There are four nat  [#permalink]

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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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26 May 2019, 15:10
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].

Veritas Prep Official Solution

Here, A, the event for which we want to find the probability is ‘each attraction gets at least one vote’.

P(A) = No of ways in which each attraction gets at least one vote /Total no. of ways in which the friends can vote.

Each attraction should get at least one vote. 6 votes can be divided among 4 attractions in the following ways: (1, 1, 1, 3) and (1, 1, 2, 2)

Case 1: (1, 1, 1, 3)

First, we select the attraction that will get 3 votes in 4 ways (= 4C1)

Now, we can select the 3 people who will vote for this attraction in 6*5*4/3! = 20 ways (= 6C3 )

The other 3 votes will be distributed among the other 3 attractions in 3! = 6 ways

The 6 people could vote for the 4 attractions in this case in 4*20*6 = 480 ways

Case 2: (1, 1, 2, 2)

Let’s select the two attractions that will get 2 votes each in 4*3/2! = 6 ways (= 4C2). Say we select caves and waterfall.

Now, we can select the 2 people who will vote for one of the selected attractions in 6*5/2! = 15 ways (= 6C2)

We can select the other 2 people who will vote for the other selected attraction in 4*3/2! = 6 ways (= 4C2)

The other 2 votes will be distributed among the other 2 attractions in 2! = 2 ways

The 6 people could vote for the 4 attractions in this case in 6*15*6*2 = 1080 ways

Total number of ways in which 6 votes can be distributed among 4 attractions such that each attraction gets at least one vote = 480 + 1080 = 1560 ways

As we saw in the questions above, the total no. of ways in which the friends can vote = 4^6

Therefore, P(A) = 1560/(4^6)
Manager
Joined: 06 Jun 2019
Posts: 140
Re: Six friends live in the city of Monrovia. There are four nat  [#permalink]

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15 Jun 2019, 06:18
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.

Hi Bunuel,

Could you please evaluate the below logic?

The ways of voting in which at most 3 attarctions participate: ABC or ABD or ACD or BCD. So overall $$4*3^6$$ cases. But those votings double count the cases consisting of only AB, AC, AD, BC, BD, and CD. Thus we need to subtract $$6*2^6$$ from $$4*3^6$$. Next, AB, AC, AD, BC, BD, and CD themselves triple count the cases consisting of only A, B, C, and D. So, we we need to add $$4$$ back to overall computation in order to include 4 cases - AAAAAA, BBBBBB, CCCCCC, DDDDDD, thus:

$$4*3^6 - 6*2^6 + 4 = 2536$$

Number of cases with at least one attraction: $$4^6 - 2536 = 1560$$

Is this logic correct?

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

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15 Jun 2019, 07:14

I did it this way:
out of 6 friends lets select 4 friends who vote for the 4 different places: 6C4 ways.
These votes can be arranged in 4! ways..
For the remaining 2 votes each has 4 ways.
So the numerator becomes: 6C4*4!*4*4....
Denominator is obviously 4^6
Please tell me where am i wrong?
Thanks
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Posts: 9869
Location: Pune, India
Re: Six friends live in the city of Monrovia. There are four nat  [#permalink]

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16 Jun 2019, 23:25
Debashis Roy wrote:

I did it this way:
out of 6 friends lets select 4 friends who vote for the 4 different places: 6C4 ways.
These votes can be arranged in 4! ways..
For the remaining 2 votes each has 4 ways.
So the numerator becomes: 6C4*4!*4*4....
Denominator is obviously 4^6
Please tell me where am i wrong?
Thanks

You need the number of distinct voting patterns for the 6 friends: F1, F2, ..., F6
The voting patterns would be like WWWWWW, WLCLCW, SLLLLL, LCLCWS etc

Using your method, you are double counting some favourable cases.
Say you select F1, F2, F3 and F5. They vote for WSLC.
F4 votes for L and F6 votes for C so you get WSLLCC.

Now imagine the selected friends are F1, F2, F3 and F6. They vote for WSLC.
F4 votes for L and F5 votes for C so you get WSLLCC again.

But since you are selecting 4 friends in different ways, you are counting them as different voting patterns which is not correct.
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Re: Six friends live in the city of Monrovia. There are four nat   [#permalink] 16 Jun 2019, 23:25
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