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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?

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.

Re: Given that there are 6 married couples. If we select only 4 [#permalink]
16 Dec 2011, 12:11

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

Re: Given that there are 6 married couples. If we select only 4 [#permalink]
08 Dec 2013, 08:53

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\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). _______________________

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 _________________

“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”

\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). _______________________

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). _________________

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!?