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Given that there are 6 married couples. If we select only 4

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Given that there are 6 married couples. If we select only 4 [#permalink] New post 18 Jan 2008, 00:05
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Given that there are 6 married couples. If we select only 4 people out of the 12, what is the probability that none of them are married to each other?
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Re: combinatorics [#permalink] New post 18 Jan 2008, 00:33
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p=\frac{12}{12}*\frac{10}{11}*\frac{8}{10}*\frac{6}{9}=\frac{16}{33}

or

p=\frac{C^6_4*(C^2_1)^4}{C^{12}_4}=\frac{16}{33}

or

p=\frac{P^6_4*(P^2_1)^4}{P^{12}_4}=\frac{16}{33}
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Last edited by walker on 19 Jan 2008, 01:53, edited 3 times in total.
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Re: combinatorics [#permalink] New post 18 Jan 2008, 15:03
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dominion wrote:
Given that there are 6 married couples. If we select only 4 people out of the 12, what is the probability that none of them are married to each other?


Prob: 1*10/11*8/10*6/9 = 16/33
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Re: combinatorics [#permalink] New post 19 Jan 2008, 01:41
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dominion wrote:
walker wrote:
p=\frac{C^6_4*(C^2_1)^4}{C^{12}_4}=\frac{16}{33}


walker, explain your combinatorics method please.


C^6_4 - we choose 4 couples of 6 ones.

C^2_1 - we chose one people of 2 ones for one couple.

(C^2_1)^4 - we have 4 couple and for each we choose one people of 2 ones for one couple.

C^{12}_4 - the total number of combinations to choose 4 people from 12 people.
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Last edited by walker on 19 Jan 2008, 01:43, edited 1 time in total.
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Re: combinatorics [#permalink] New post 19 Jan 2008, 01:52
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p=\frac{12}{12}*\frac{10}{11}*\frac{8}{10}*\frac{6}{9}=\frac{16}{33}

\frac{12}{12} - we choose 12 of 12.

\frac{10}{11} - we choose 10=12-1(prevous choice)-1(another people out of couple) of 11=12-1(prevous choice).

\frac{8}{10} - we choose 8=12-2(prevous choice)-2(another people out of couple) of 10=12-2(prevous choice).

\frac{6}{9} - we choose 6=12-3(prevous choice)-3(another people out of couple) of 9=12-3(prevous choice).

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p=\frac{P^6_4*(P^2_1)^4}{P^{12}_4}=\frac{16}{33}
the same logic as for C_m^n
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Re: combinatorics [#permalink] New post 24 Aug 2008, 13:46
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dominion wrote:
Given that there are 6 married couples. If we select only 4 people out of the 12, what is the probability that none of them are married to each other?


= (12C1 * 10C1 * 8C1*6C1)/4!/ 12C4
= 12*10*8*6 / (12*11*10*9) = 16/33
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Re: combinatorics [#permalink] New post 05 Feb 2009, 17:36
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C^6_4 - we choose 4 couples (not people) out of 6 couples (the number of all couples)

the next step: we choose one person out of each couple - C^2_1
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Re: combinatorics [#permalink] New post 05 Feb 2009, 19:00
walker wrote:
C^6_4 - we choose 4 couples (not people) out of 6 couples (the number of all couples)

the next step: we choose one person out of each couple - C^2_1


Walker,
Ok, its making better sense now.
Are we choosing 4 couples because that means "not 4 people", ie 12-4=8?
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Re: combinatorics [#permalink] New post 05 Feb 2009, 23:50
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dominion wrote:
Given that there are 6 married couples. If we select only 4 people out of the 12, what is the probability that none of them are married to each other?


So, we should count how many possibilities we have to form group of 4 people with restriction: none of them are married to each other.
There a few ways to count all combinations. One of the ways is following: 4 people belong to 4 distinct couples. Therefore, we could choose these 4 couples and then 1 person out of each couple - C^6_4*(C^2_1)^4

look at other problem:
Quote:
Given that there are 8 soccer teams. If we select only 6 people out of the 88 (8 teams, 11 people in each team), what is the probability that none of them are out of the same team?


we can use the same reasoning: choose 6 teams out of 8 teams (our 6 people are from 6 different teams) and then choose 1 player out of 11 for each team.

I hope it is clearer now.
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Re: combinatorics [#permalink] New post 06 Feb 2009, 09:22
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dominion wrote:
Given that there are 6 married couples. If we select only 4 people out of the 12, what is the probability that none of them are married to each other?

total ways of choosing 4 out of 12:12x11x10x9
first person can be chosen in 12 ways. The next person can be chosen in 10 ways (because we don't want spouses to be in the group). The next person can be again chosen in only 8 ways (out of the 10 people left, we have to exclude 2 whose spouses we have already selected). And the last person in 6 ways.

hence, probability=12x10x8x6/(12x11x10x9) = 16/33.
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Re: combinatorics [#permalink] New post 06 Feb 2009, 18:06
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botirvoy wrote:
x2suresh wrote:
dominion wrote:
Given that there are 6 married couples. If we select only 4 people out of the 12, what is the probability that none of them are married to each other?


= (12C1 * 10C1 * 8C1*6C1)/4!/ 12C4
= 12*10*8*6 / (12*11*10*9) = 16/33

suresh, can you please explain the logic of this calculation?
Thank you!


12C1 = select any person from 6 married couples (12 person)
10C1 = select second person from remaining people and exclude the first person's spouse
8C1 = select 3rd person from remain people exclue first and second perssons's spouses
6C1 = select 4th person from remaining people exclude 1st,2nd ,3rd persons's spouse

Becuase order is not matter.. you need to divide by 4!

Did you get it?

=
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Re: combinatorics [#permalink] New post 23 Mar 2009, 10:20
xALIx wrote:
walker wrote:
dominion wrote:
Given that there are 6 married couples. If we select only 4 people out of the 12, what is the probability that none of them are married to each other?


So, we should count how many possibilities we have to form group of 4 people with restriction: none of them are married to each other.
There a few ways to count all combinations. One of the ways is following: 4 people belong to 4 distinct couples. Therefore, we could choose these 4 couples and then 1 person out of each couple - C^6_4*(C^2_1)^4

look at other problem:
Quote:
Given that there are 8 soccer teams. If we select only 6 people out of the 88 (8 teams, 11 people in each team), what is the probability that none of them are out of the same team?


we can use the same reasoning: choose 6 teams out of 8 teams (our 6 people are from 6 different teams) and then choose 1 player out of 11 for each team.

I hope it is clearer now.


Wow, makes very clear sense now. Much obliged.
Ali



another way of looking at this problem :

total number of ways = 12C4 = 495

now let's try to find out total number of unfavourable ways. We can subtract those from 495 to get the number of favourable ways.

number of ways of choosing 4 people such that there are two couples = 6C2 = 15

number of ways of choosing 4 people such that there is only one couple = number of ways of choosing one couple * number of ways of choosing two people who are not couples = 6C1 (10C2 -5) = 240.

total number of unfavourable ways = 240 + 15 = 255

total number of favourable ways = 495 - 255 = 240

required probability = 240/495 = 16/33
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Re: combinatorics [#permalink] New post 27 Sep 2009, 05:13
Given that there are 6 married couples. If we select only 4 people out of the 12, what is the probability that none of them are married to each other?


Soln:
= (12/12) * (10/11) * (8/10) * (6/9)
= 16/33
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Re: combinatorics [#permalink] New post 31 Mar 2010, 12:17
hi walker regarding the soccer question the answer should be

8C6 * (11C1)^6
---------------
88C4

right na thks

but just want to know if i have understood the concept thks
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Re: combinatorics [#permalink] New post 03 Apr 2010, 05:17
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bnagdev282 wrote:
hi walker regarding the soccer question the answer should be

8C6 * (11C1)^6
---------------
88C4

right na thks

but just want to know if i have understood the concept thks


Sorry, where did you get 88C4? It's a huge number!
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Re: combinatorics [#permalink] New post 03 Apr 2010, 14:22
hi walker i calculated it as 8 team with 11 players each so in all 88 players and chosing 4 out of them so 88c4
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Re: combinatorics [#permalink] New post 03 Apr 2010, 14:46
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bnagdev282 wrote:
hi walker i calculated it as 8 team with 11 players each so in all 88 players and chosing 4 out of them so 88c4


I see... You are right if we use 88C6 rather than 88C4 (we need 6 people). I think it is a typo, right?
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Re: combinatorics [#permalink] New post 03 Apr 2010, 14:48
ya typo error sorry

thks walker
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Re: combinatorics [#permalink] New post 11 Apr 2011, 06:43
In x2suresh's solution, I'm unable to visualize the exact arrangements because of which we should divide 12x10x8x6 ways by 4!. Could someone please explain in more detail ? The thing is, we're choosing here, so how does the question of arrangements come in picture at all ?

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Re: combinatorics [#permalink] New post 30 Apr 2011, 19:14
1 * 10/11 * 8/10 * 6/9
= 16/33
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Re: combinatorics   [#permalink] 30 Apr 2011, 19:14
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