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Re M0205 [#permalink]
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16 Sep 2014, 00:17
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Official Solution:There are four distinct pairs of brothers and sisters. In how many ways can a committee of 3 be formed and NOT have siblings in it? A. 8 B. 24 C. 32 D. 56 E. 80 Since the committee shouldn't have siblings in it, then each pair can send only one "representative" to the committee. # of ways to choose which 3 pairs of brothers and sisters will send one "representative" to the committee is \(C^3_4\) (choosing 3 pairs which will be granted the right to send one "representative" to the committee). But each of these 3 pairs can send 2 persons to the committee either a brother or a sister: \(2*2*2=2^3\). So total # of ways to form the committee is \(C^3_4*2^3=32\). Answer: C
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Since we have to select 3 with NOT a sibling pair, then:
Total no of possible ways  number of ways of selecting 3 including a sibling pair
8C3  4C1 * 6C1
(No. of ways of selecting 8 taking 3 at a time)  Number of ways of selecting a pair out of 4, multiplied with selecting the remaining one person out of the left 3 pairs (6 people)
Therefore: 56  (4*6) = 5624 = 32



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Re: M0205 [#permalink]
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20 Mar 2015, 19:53
Hi Bunuel
Can you further explain why you multiply by 2^3?
I understand that you want to understand the total number of combinations of choosing 3 people from 4 pairs (so 4C3) and then we multiply that by 2^3 because for each of the three pairs they could either send the brother or sister  thus there's two combinations? Just want to make sure I'm thinking about this correctly.



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Re: M0205 [#permalink]
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01 Apr 2015, 08:10
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Bunuel wrote: There are four distinct pairs of brothers and sisters. In how many ways can a committee of 3 be formed and NOT have siblings in it?
A. 8 B. 24 C. 32 D. 56 E. 80 Total # of committees = \(^8C_3\) = 56 Committees with 1 sibling = \(^4C_1 * ^6C_1\) = 24 56 24 = 32 Answer C
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Re: M0205 [#permalink]
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06 Apr 2015, 11:47
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Is this a possible correct solution as well ?
8*6*4  3!
Thanks you



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Re: M0205 [#permalink]
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15 Jun 2015, 06:47
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Hi Bunuel, Can you please suggest if the following approach is correct? There are 4 sibling pairs (B1,S1) (B2,S2) (B3,S3) (B4,S4) Total Number of candidates  8. For filling the first slot in the committee, we have 8 people eligible. For filling the second slot in the committee, we have 6 people eligible. [Removing one of the sibling of the earlier selected candidate] For filling the third slot in the committee, we have 4 people eligible. [Removing another of the sibling of the earlier selected candidate] Therefore total possibilities  8*6*4. However my answer does not figure in the options. Where am I going wrong? Vijay Bunuel wrote: Official Solution:
There are four distinct pairs of brothers and sisters. In how many ways can a committee of 3 be formed and NOT have siblings in it?
A. 8 B. 24 C. 32 D. 56 E. 80
Since the committee shouldn't have siblings in it, then each pair can send only one "representative" to the committee. # of ways to choose which 3 pairs of brothers and sisters will send one "representative" to the committee is \(C^3_4\) (choosing 3 pairs which will be granted the right to send one "representative" to the committee). But each of these 3 pairs can send 2 persons to the committee either a brother or a sister: \(2*2*2=2^3\). So total # of ways to form the committee is \(C^3_4*2^3=32\).
Answer: C



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Re: M0205 [#permalink]
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15 Jun 2015, 06:52
svijayaug12 wrote: Hi Bunuel, Can you please suggest if the following approach is correct? There are 4 sibling pairs (B1,S1) (B2,S2) (B3,S3) (B4,S4) Total Number of candidates  8. For filling the first slot in the committee, we have 8 people eligible. For filling the second slot in the committee, we have 6 people eligible. [Removing one of the sibling of the earlier selected candidate] For filling the third slot in the committee, we have 4 people eligible. [Removing another of the sibling of the earlier selected candidate] Therefore total possibilities  8*6*4. However my answer does not figure in the options. Where am I going wrong? Vijay Bunuel wrote: Official Solution:
There are four distinct pairs of brothers and sisters. In how many ways can a committee of 3 be formed and NOT have siblings in it?
A. 8 B. 24 C. 32 D. 56 E. 80
Since the committee shouldn't have siblings in it, then each pair can send only one "representative" to the committee. # of ways to choose which 3 pairs of brothers and sisters will send one "representative" to the committee is \(C^3_4\) (choosing 3 pairs which will be granted the right to send one "representative" to the committee). But each of these 3 pairs can send 2 persons to the committee either a brother or a sister: \(2*2*2=2^3\). So total # of ways to form the committee is \(C^3_4*2^3=32\).
Answer: C Check here: iftherearefourdistinctpairsofbrothersandsisters99992.html#p775925Hope it helps.
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Re: M0205 [#permalink]
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07 Nov 2015, 04:49
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[quote="svijayaug12"]Hi Bunuel,
Can you please suggest if the following approach is correct?
There are 4 sibling pairs
(B1,S1) (B2,S2) (B3,S3) (B4,S4)
Total Number of candidates  8.
For filling the first slot in the committee, we have 8 people eligible. For filling the second slot in the committee, we have 6 people eligible. [Removing one of the sibling of the earlier selected candidate] For filling the third slot in the committee, we have 4 people eligible. [Removing another of the sibling of the earlier selected candidate]
Therefore total possibilities  8*6*4.
However my answer does not figure in the options. Where am I going wrong?
Vijay
Vijay, Your approach is correct not complete . Since this is a selection problem, order of selection is not important. In your approach , you may have a committee with B1,S2,B3 as well as S2,B1,B3 .. if you notice this is due to the number of ways each is permitted. To eliminate this , you need to divide the result by 3!. Hope this helps.
Sunil



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Re: M0205 [#permalink]
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26 Nov 2015, 17:40
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I attacked this question in a different way: we have 8 people, out of which we have to choose 3. Thus, we have 8C3 ways to do so: 8C3 = 8*7*6/3! = 56. Note that there are way fewer ways to choose in such ways that no brothers are selected. We can eliminate D and E. now, look at B and C. B+C = 56, clearly, the 2 options are complementary, and there is a high chance that the answer is one of the two. A is thus out. It's a good way to make a strategic guess, chances to answer correctly  50% now, in how many ways can we select 3 people so that no brothers/sisters are in the committee? well, let's see how many committees, in which the restriction is not respected, we can get! now, there are 4 brothers and 4 sisters! suppose we select 1 brother and 1 sister > we then have 1*4C1*6C1 or 4*6= 24 ways, we're almost there! why 4C1  from 4, we have to choose 1 that will be brother/sister of the other one. now, we have chosen 2 members, out of 6, we have to choose 1. that's why 6C1 alternatively, we can assign letters and try to list all the combinations, I know it is not preferable, but still 24  not that much! A1B1 A2B2 A3B3 A4B4 now, we can have A1B1*6C1; A2B2*6C1; A3B3*6C1; A4B4*6C1 > clearly we have 4*6C1!!! It doesn't matter the order A1B1 ir B1A1:) now, the final step: 56 ways total  24 ways in which the restriction is not respected = 32 ways in which restriction is respected.



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Another solution:
1st person: There are 8 ways of selection.
2nd person: Remaining 7 people, but not choosing the person who is pair matching with 1st, so there are 6 ways of selection.
3rd person: Remaining 6 people, not choosing the people who are pair maching with 1st and 2nd, so there are 4 ways of selection.
In addition, choosing no order, divided 3!
The no. of ways = 8*6*4/3! = 32 ways.



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Re M0205 [#permalink]
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12 May 2016, 07:53
4 pairs means 8 right? Could this question be explained in a better way?



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Re: M0205 [#permalink]
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12 May 2016, 15:11



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Re: M0205 [#permalink]
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24 Jul 2016, 08:09
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I found the following solution, bit lengthy but worked for me Four pairs means  4B and 4S we can select 1B out of 4B > 4C1 corresponding to that I have to select 2S out of 3S (because i have to exclude 1S corresponding to 1B, which are making pair) > 3C2 So for the first combination it would be 4C1 * 3C2
B,S > (1,2); (2,1); (3,0); (0,3)
The rest of the pairs would be (4C2*2C1); (4C3*1); (1*4C3)
So total required combination is 4C1*3C2 + 4C2*2C1 + 4C3*1 + 1*4C3 = 12+12+4+4=32



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Re: M0205 [#permalink]
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24 Aug 2016, 09:46
(B1,S1), (B2,S2), (B3,S3), (B4,S4) = 04 PAIR, TOTAL 08 PERSON
so total combination of 03 person 8c3= 56 LETS SAY NOW (B1,S1) one pair, so we can choose the third place from 06 person now, means when first borther and sisiter are always the part of the combination third spot can be choosen by any of the remaining 06
ii) if (b2, s2) pair is choosen  so rest from 06, so total 06 combination
iii) (b3,s3) = 06
(iv) (b4,s4)= 06
so total combination when pair is always there = 06*04= 24
so total without sibling = 5624= 32



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Re: M0205 [#permalink]
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19 Feb 2017, 12:18
I went through this problem in CAT: I broke the question in this way: 3 possibilities: 3 man or 3 girls or (2 man and 1 girl) or (2 girls and 1 man) Since or means add and we have 2 things that are ''simmetric'' combo of 3 man=combo of 3 girls=C4,3=4 combo of (2 man and 1 girl)=combo of (2 girls and a man) here the calculation for me was not immediate i did it this way: the combo of 2 man are C4,, now for each combo of man you can have 2 possible girls(the ones that are not sisters) so all the combo of 2 men and 1 girl( that as said are equal to the combo of 2 girls and one man) are:2C4,2 suming all up: 2*C4,3+2*2*C4,2=32



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I approached it this way: Suppose these are the pairs: ab cd ef gh Now to select three ppl from 4 pairs, as per the image Total = 4*8 = 32
>> !!!
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Re: M0205 [#permalink]
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12 Sep 2017, 01:53
Easiest way to approach this is restriction method  Total number of combination that can be formed from 8 people  8C3 Number of groups possible which includes atleast a brother sister pair and a person from remaining  4C1 ( choosing 1 out of 4 pairs of brother & sister) * 6C1 Final answer  8C3  (4C1*6C1) = 32



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Re: M0205 [#permalink]
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14 Oct 2017, 16:11
Please help understand where my thinking is flawed ? A 3 member committee needs to be formed from 4 pair of distinct siblings say (b1,s1), (b2, s2), (b3, s3), (b4,s4).
1st member can be any of the 8 persons. Now, for next member selection, discard the pair from which the 1st member was selected. So now we have 3 pairs left to choose from. 2nd member can be any of the 6 persons in the left 3 pairs. Now, for next member selection, discard the pair from which the 1st and 2nd member were selected. So now we have 2 pairs left to choose from. 3rd member can be any of the 4 persons in the remaining 2 pairs.
So total number of ways = 8*6*4 = 192. Where am I counting the extras ?



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Re: M0205 [#permalink]
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15 Oct 2017, 01:57
shohm wrote: Please help understand where my thinking is flawed ? A 3 member committee needs to be formed from 4 pair of distinct siblings say (b1,s1), (b2, s2), (b3, s3), (b4,s4).
1st member can be any of the 8 persons. Now, for next member selection, discard the pair from which the 1st member was selected. So now we have 3 pairs left to choose from. 2nd member can be any of the 6 persons in the left 3 pairs. Now, for next member selection, discard the pair from which the 1st and 2nd member were selected. So now we have 2 pairs left to choose from. 3rd member can be any of the 4 persons in the remaining 2 pairs.
So total number of ways = 8*6*4 = 192. Where am I counting the extras ? Check here: http://gmatclub.com/forum/ifthereare ... ml#p775925Hope it helps.
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