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Re: combinatorics [#permalink]
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02 May 2011, 19:44
tusharvk 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? 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. hi, why will total ways of choosing 4 out of 12:12x11x10x9 not be 12C4 ? I understand the logic of 12*11*10*9 but I am not able to convince myself why its not 12C4. Please explain. Thanks.



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Re: combinatorics [#permalink]
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11 Sep 2011, 23:31
I dont get why we are dividing by 4!...because order doesnt matter therefore we are using C and not P...then why use 4! seofah 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!



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Re: Given that there are 6 married couples. If we select only 4 [#permalink]
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16 Dec 2011, 13:11
Hey please suggest where I am going wrong.
probability that none of the couple is on the group is = 1(probability that exactly 1 couple is there)  (probability that 2 couples are there
probability of 1 couple being there : of the 6 couples an 1 can be selected in 6C1 = 6 ways . the third person can be chosen in 10 ways (leaving out the couple selected). the 4th person can be selected in 8 ways (leaving out the couple selected, the 3rd person selected and the spouse of the 3rd person selected) = 6*10*8 = 480 ways.. since total number of ways = 12C4. therefore probability that 1 couple is part of teh selection is 480/12C4
probability of both couple being there = 6C2/12C4.
therefore probability that no couple is there : 1 [(480/12C4)+(6C2/12C4)] = 1/2
Please tell me what is wrong with this method.. Many Thanks



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Re: Given that there are 6 married couples. If we select only 4 [#permalink]
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08 Dec 2013, 10:45
Quote: hi,
why will total ways of choosing 4 out of 12:12x11x10x9 not be 12C4 ?
I understand the logic of 12*11*10*9 but I am not able to convince myself why its not 12C4.
Please explain. Thanks I got the same doubts. Please help!!



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Re: Given that there are 6 married couples. If we select only 4 [#permalink]
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09 Dec 2013, 00:56



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Re: Given that there are 6 married couples. If we select only 4 [#permalink]
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09 Dec 2013, 01:28
Quote: Bunuel wrote: Quote:
Yes, the total number of ways to choose 4 people out of 12 is C^4_{12}=\frac{12!}{8!4!}. But if I do: (12C1*10C1*8C1*6C1)/12C4 the answer is not 16/33 What am I missing here??



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Re: Given that there are 6 married couples. If we select only 4 [#permalink]
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09 Dec 2013, 01:45



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Re: combinatorics [#permalink]
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09 Feb 2014, 02:08
walker wrote: \(p=\frac{12}{12}*\frac{10}{11}*\frac{8}{10}*\frac{6}{9}=\frac{16}{33}\)
\(\frac{12}{12}\)  we choose 12 of 12.
\(\frac{10}{11}\)  we choose 10=121(prevous choice)1(another people out of couple) of 11=121(prevous choice).
\(\frac{8}{10}\)  we choose 8=122(prevous choice)2(another people out of couple) of 10=122(prevous choice).
\(\frac{6}{9}\)  we choose 6=123(prevous choice)3(another people out of couple) of 9=123(prevous choice). _______________________
\(p=\frac{P^6_4*(P^2_1)^4}{P^{12}_4}=\frac{16}{33}\) the same logic as for \(C_m^n\) The Question was "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?"I am not able to understand why "\(\frac{12}{12}\)  we choose 12 of 12."Shouldn't this be 1/12 because the first person can be any 1 of the 12......There is something I am missing..... Math Gurus: Your inputs please
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Re: combinatorics [#permalink]
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09 Feb 2014, 02:15
WoundedTiger wrote: walker wrote: \(p=\frac{12}{12}*\frac{10}{11}*\frac{8}{10}*\frac{6}{9}=\frac{16}{33}\)
\(\frac{12}{12}\)  we choose 12 of 12.
\(\frac{10}{11}\)  we choose 10=121(prevous choice)1(another people out of couple) of 11=121(prevous choice).
\(\frac{8}{10}\)  we choose 8=122(prevous choice)2(another people out of couple) of 10=122(prevous choice).
\(\frac{6}{9}\)  we choose 6=123(prevous choice)3(another people out of couple) of 9=123(prevous choice). _______________________
\(p=\frac{P^6_4*(P^2_1)^4}{P^{12}_4}=\frac{16}{33}\) the same logic as for \(C_m^n\) The Question was "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?"I am not able to understand why "\(\frac{12}{12}\)  we choose 12 of 12."Shouldn't this be 1/12 because the first person can be any 1 of the 12......There is something I am missing..... Math Gurus: Your inputs please 12/12=1 is another way of saying: choosing ANY for the first person (probability of 1).
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Re: combinatorics [#permalink]
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20 Apr 2014, 13:48
walker wrote: \(p=\frac{12}{12}*\frac{10}{11}*\frac{8}{10}*\frac{6}{9}=\frac{16}{33}\)
or
\(p=\frac{C^6_4*(C^2_1)^4}{C^{12}_4}=\frac{16}{33}\)
or
\(p=\frac{P^6_4*(P^2_1)^4}{P^{12}_4}=\frac{16}{33}\) Hi  can someone please explain the reasoning for the last two equations? In the second equation, it stating the different ways 4 couples out of six can be selected(to give 4 individual people), followed by how many different ways to select 1 person out of the 2 people in a couple(and do it 4 times) and follow that by dividing the whole quantity by 12C4. Can you explain WHY you're doing this?Same with the third equation  what's different here now that we have 4P6 and not 6C4? An additional question  I've seen the equation (12C1 * 10C1 * 8C1*6C1)/4!/ 12C4  why do we divide by 4! instead of multiply by 4!?



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Re: Given that there are 6 married couples. If we select only 4 [#permalink]
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02 Aug 2014, 07:15
Bunuel wrote: nikhil09 wrote: Quote: Bunuel wrote: Quote:
Yes, the total number of ways to choose 4 people out of 12 is C^4_{12}=\frac{12!}{8!4!}. But if I do: (12C1*10C1*8C1*6C1)/12C4 the answer is not 16/33 What am I missing here?? That's because if you do this way it's not correct. Yes i guess that is because we have to divide (12C1*10C1*8C1*6C1) by 4!, but i don't really get why Is that because the order does not matter?



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Re: Given that there are 6 married couples. If we select only 4 [#permalink]
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24 Sep 2015, 07:43
walker wrote: \(p=\frac{12}{12}*\frac{10}{11}*\frac{8}{10}*\frac{6}{9}=\frac{16}{33}\)
or
\(p=\frac{C^6_4*(C^2_1)^4}{C^{12}_4}=\frac{16}{33}\)
or
\(p=\frac{P^6_4*(P^2_1)^4}{P^{12}_4}=\frac{16}{33}\) Hey walker, Can you tell me why there is no difference between the results of combination and permutation formula. Hope I am not asking something odd.
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Re: Given that there are 6 married couples. If we select only 4 [#permalink]
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