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If 4 people are selected from a group of 6 married couples
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13 Aug 2010, 07:42
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If 4 people are selected from a group of 6 married couples, what is the probability that none of them would be married to each other? A. 1/33 B. 2/33 C. 1/3 D. 16/33 E. 11/12
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Re: confuseddd
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13 Aug 2010, 08:10
bibha wrote: If 4 people are selected from a group of 6 married couples, what is the probability that none of them would be married to each other? 1/33 2/33 1/3 16/33 11/12 Each couple can send only one "representative" to the committee. We can choose 4 couples (as there should be 4 members) to send only one "representatives" to the committee in \(C^4_6\) # of ways. But these 4 chosen couples can send two persons (either husband or wife): \(2*2*2*2=2^4\). So # of ways to choose 4 people out 6 married couples so that none of them would be married to each other is: \(C^4_6*2^4\). Total # of ways to choose 4 people out of 12 is \(C^4_{12}\). \(P=\frac{C^4_6*2^4}{C^4_{12}}=\frac{16}{33}\) Answer: D. Similar problems with different approaches: combinationpermutationproblemcouples98533.html?hilit=married%20couplespscombinations94068.html?hilit=married%20couplescommitteeof88772.html?hilit=married%20couplesHope it helps.
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If 4 people are selected from a group of 6 married couples
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19 Feb 2012, 08:48
+1 D A faster way to solve it: \(\frac{12}{12} * \frac{10}{11} * \frac{8}{10} * \frac{6}{9} = \frac{48}{99} = \frac{16}{33}\)
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Re: confuseddd
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13 Sep 2010, 20:20
harithakishore wrote: But these 4 chosen couples can send two persons (either husband or wife): 2*2*2*2
Bunuel,can you please..please explain this..im confused.... 4 chosen couples...i think we can choose 4 different people and not couples..im really confused and also how come it is 2*2*2*2...please explain We have 6 couples: \(A (a_1, a_2)\); \(B (b_1, b_2)\); \(C (c_1, c_2)\); \(D (d_1, d_2)\); \(E (e_1, e_2)\); \(F (f_1, f_2)\); We should choose 4 people so that none of them will be married to each other. The above means that 4 chosen people will be from 4 different couples, for example from A, B, C, D or from A, D, E, F... The # of ways to choose from which 4 couples these 4 people will be is \(C^4_6=15\); Let's consider one particular group of 4 couples: {A, B, C, D}. Now, from couple A in the group could be either \(a_1\) or \(a_2\), from couple B in the group could be either \(b_1\) or \(b_2\), from couple C in the group could be either \(c_1\) or \(c_2\), and from couple D in the group could be either \(d_1\) or \(d_2\). So each couple has two options (each couple can be represented in the group of 4 people by \(x_1\) or \(x_2\)), so one particular group of 4 couples {A, B, C, D} can give us \(2*2*2*2=2^4\) groups of 4 people from different couples. One particular group of 4 couples {A, B, C, D} gives \(2^4\) groups of 4 people from different couples > 15 groups give \(15*2^4\) groups of 4 people from different couples (total # of ways to choose 4 people so that no two will be from the same couple) . Hope it's clear.
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Re: confuseddd
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14 Aug 2010, 07:27
bibha wrote: Why can't we do it like this:
total ways of selective 4 ppl from 6 married couples = 12C4 Favorable outcome = 12 *10*8*6 ???? The way you are doing is wrong because 12*10*8*6=5760 will contain duplication and if you are doing this way then to get rid of them you should divide this number by the factorial of the # of people  4! > \(\frac{5760}{4!}=240=C^2_4*2^8=favorable \ outcomes\). Consider this: there are two couples and we want to choose 2 people not married to each other. Couples: \(A_1\), \(A_2\) and \(B_1\), \(B_2\). Committees possible: \(A_1,B_1\); \(A_1,B_2\); \(A_2,B_1\); \(A_2,B_2\). Only 4 such committees are possible. If we do the way you are doing we'll get: 4*2=8. And to get the right answer we should divide 8 by 2! > 8/2!=4. Hope it helps.
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Re: If 4 people are selected from a group of 6 married couples
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09 Mar 2019, 19:00
j.shivank wrote: Dear Bunuel chetan2uWhat if we first select 1 individual from each set of couple and then out of those individuals, select 4 individuals. (2C1 * 2C1 * 2C1 * 2C1 * 2C1 * 2C1) * 6C4 = 960 Then, Total no. of ways to choose 4 people out of 12 is = 12C4 = 495 But turns out that this method is wrong as the probability comes out to be greater than 1 Can please help in pointing out why selecting 6 individuals from 6 couples first, is wrong? There are repetitions in this.. One example.. Say 12, 34,56,78,910,1112 are pair.. Select one out of each 1,3,5,7,9,11 choose ways of choosing 4: two ways would be 1,3,5,7 or 1,3,5,9 Now, choose one as 1,3,5,7,9,12 4: two ways would be 1,3,5,7 or 1,3,5,9 Not only these two, there will be many repetitions.. That is why you get probability more than 1.. But if you take 12 people first, remove the partner of the one you have chosen .. So 10 left, close one , and again remove the partner.. Now you are left with 8... And so on
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Re: If 4 people are selected from a group of 6 married couples
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24 Apr 2014, 23:02
russ9 wrote: JusTLucK04 wrote: Also Bunuel...One of my pain points in PnC is when the number of things to be allocated is more than the number of people...Say I have 100 pencils to be distributed 10 students..Such that each can get anything between 1 to 100.What is the formulaic approach to such questions? I would let Bunuel answer this but my thought would be: treat it as equal distribution and 100c10? I think it should be...100*99....91*90 And if the question mentions that it is possible that a student recieves not even a single pencil..I think we go case wise with 1 student gets all..2 student get all pencils..and so on



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Re: If 4 people are selected from a group of 6 married couples
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15 Dec 2014, 06:21
Bunnel, Whats wrong with this approach ? Case1 12c46c2 ( Both couples selected) Case2 12c4 6c1 * 10c1 * 8c1( One couple & 2 Noncouple )
Sum up the putcomes of both the cases & find probability.
Please tell how to approach this



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Re: If 4 people are selected from a group of 6 married couples
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16 Dec 2017, 08:41
Erjan_S wrote: metallicafan wrote: +1 D
A faster way to solve it:
\(\frac{12}{12} * \frac{10}{11} * \frac{8}{10} * \frac{6}{9} = \frac{48}{99} = \frac{16}{33}\) Isn't it a short and great approach? is there anything wrong with it? Why for problems like this is combinatorics a preferred approach? Hi Erjan_SThere is nothing wrong with the approach. As the numbers are simple so you can simply count the possible and total outcomes and then keep on removing the pairs. There could be multiple ways to solve counting problems. P & C is preferred because you get the value through direct formula. in case of simple numbers you can manually count the number of possible ways or selections, but when the numbers or situation become a bit complex then you will have to rely upon P & C.



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Re: confuseddd
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14 Aug 2010, 06:53
Why can't we do it like this:
total ways of selective 4 ppl from 6 married couples = 12C4 Favorable outcome = 12 *10*8*6 ????



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Re: confuseddd
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16 Aug 2010, 11:36
awesome explanation +1



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Re: confuseddd
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13 Sep 2010, 19:38
But these 4 chosen couples can send two persons (either husband or wife): 2*2*2*2
Bunuel,can you please..please explain this..im confused.... 4 chosen couples...i think we can choose 4 different people and not couples..im really confused and also how come it is 2*2*2*2...please explain



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Re: confuseddd
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13 Sep 2010, 20:27
Thats a fantabulous explanation...thankyou so much....



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Re: If 4 people are selected from a group of 6 married couples
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19 Jan 2012, 00:13
Total possible selection = 12!/(4!*8!)= 11*45 (after simplification) Favourble out come can be obtained by the multiplying the following combinations. 1. We require only 4 people. So these 4 are going to be from 4 different groups. Total availbale grops =6. So this combination is 6c4 = 6!/(4!*2!) =15 2. Select 1 member from each group = 2c1*2c1*2c1*2c1=2^4=16
Probability = (15*16)/(11*45)=16/33



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Re: If 4 people are selected from a group of 6 married couples
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27 Dec 2012, 18:34
bibha wrote: If 4 people are selected from a group of 6 married couples, what is the probability that none of them would be married to each other? 1/33 2/33 1/3 16/33 11/12 If we are to select 4 people from 6 couples WITHOUT any restriction, how many ways can we make the selection? 12!/4!6! = 11*5*9 = 495 If we are to select 4 people from 6 couples WITH restriction that no married couple can both make it to the group, only a representative? 6!/4!2! = 15 But we know that to select a person from each couple, take 2 possibilities15*2*2*2*2 = 240 Probability = Desired/All Possibilities = 240/495 = 16/33 Answer: D



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Re: If 4 people are selected from a group of 6 married couples
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23 Apr 2014, 19:42
metallicafan wrote: +1 D
A faster way to solve it:
\(\frac{12}{12} * \frac{10}{11} * \frac{8}{10} * \frac{6}{9} = \frac{48}{99} = \frac{16}{33}\) I believe that I had seen elsewhere that IF we were doing this same problem without the probability part of the question, we would have to divide (12x10x8x6) with 4!. Why is that not applicable when doing probability? Don't we still need the favorable outcomes?



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Re: If 4 people are selected from a group of 6 married couples
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23 Apr 2014, 22:28
Number of ways to select...Atleast one married couple is
6C1 Choose 1 married couple for 2 seats 10C2 Chose 2 members from the remaining 5 couples for the remaining 2 seats
6C1*10C2= 270 No of ways to select 4 people from 12=12C4
P= 270/495
Answer=1 the prob of the above =5/11
What am I missing here?



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Re: If 4 people are selected from a group of 6 married couples
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Updated on: 24 Apr 2014, 22:34
Also Bunuel...One of my pain points in PnC is when the number of things to be allocated is more than the number of people...Say I have 100 pencils to be distributed 10 students..Such that each can get anything between 1 to 100.What is the formulaic approach to such questions?
Originally posted by JusTLucK04 on 23 Apr 2014, 22:49.
Last edited by JusTLucK04 on 24 Apr 2014, 22:34, edited 1 time in total.



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Re: If 4 people are selected from a group of 6 married couples
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24 Apr 2014, 18:20
JusTLucK04 wrote: Also Bunuel...One of my pain points in PnC is when the number of things to be allocated is more than the number of people...Say I have 100 pencils to be distributed 10 students..Such that each can get anything between 1 to 100.What is the formulaic approach to such questions? I would let Bunuel answer this but my thought would be: treat it as equal distribution and 100c10?



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Re: If 4 people are selected from a group of 6 married couples
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26 Apr 2014, 01:38
1. Select the 1st person: sure there is 1st'spouse in the group > then 11 left 2. Select the 2nd person: probability not choose 2rd's spouse is 10/11 > then 10 left (2 ppl are 1st and 2nd's spouses) 3. Select the 3rd person: probability 8/10 > then 9 left (3ppl are 1st, 2nd and 3rd spouses) 4. Select the 4th: probability 6/9 > Probability when choose 4 ppl = 10/11*8/10*6/9 = 11/33




Re: If 4 people are selected from a group of 6 married couples
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26 Apr 2014, 01:38



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