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If there are four distinct pairs of brothers and sisters [#permalink]
 28 Aug 2010, 17:14
Question Stats:
64% (01:29) correct
35% (00:41) wrong based on 42 sessions
If there are four distinct pairs of brothers and sisters, then 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. 192
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Re: If there are four distinct pairs of brothers and sisters [#permalink]
28 Aug 2010, 17:32
seekmba wrote: If there are four distinct pairs of brothers and sisters, then in how many ways can a committee of 3 be formed and NOT have siblings in it?
8
24
32
56
192
I find it difficult to understand the difference between permutation and combination and hence find these questions very hard. As committee shouldn't have siblings in it, then a pair can send only one "representative" to the committee. # of ways to choose which 3 pairs of brothers and sisters should 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 is C^3_4*2^3=32. Answer: C. Similar problems: confuseddd-99055.html?hilit=marriedps-combinations-94068.html?hilit=marriedcombination-permutation-problem-couples-98533.html?hilit=marriedHope it helps.
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Re: If there are four distinct pairs of brothers and sisters [#permalink]
01 Sep 2010, 00:28
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let AABBCCDD is the group with same letter representing a sibling pair.
total # ways to select 3 from 8 is 8C3=56
Qtn: committee of 3 NOT having siblings in it = Total (56) - committee of 3 with siblings in it
committee of 3 with siblings = select 2As and one from rem. 6. This can be done in 4 ways as 4 different letters A,B,C and D
= 4 * (2C2*6C1) = 24
Hence Answer = 56-24 = 32
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Re: If there are four distinct pairs of brothers and sisters [#permalink]
06 Sep 2010, 12:29
The first person on the committee can be anyone of the 8.
The second person on the committee can be only one out of 6 (the first person with her or his sibling excluded).
The third person can be selected only out of 4 (first two members and their siblings excluded), giving 8*6*4 permutations. These are called permutations because we count every distinct combination of 3 people 6 times (6=3!). For example, we count {Ann, Bill, and Jane} as {Ann, Bill, Jane}, {Ann, Jane, Bill}, {Jane, Ann, Bill}, {Jane, Bill, Ann}, {Bill, Ann, Jane}, {Bill, Jane, Ann}. To get the number of combinations, divide the number of permutations by 6: 8*6*4 / (6) = 32 combinations.
I understand uptil the last part but I don't understand why we're diving by 6? Can anyone clarify this for me?
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Re: If there are four distinct pairs of brothers and sisters [#permalink]
06 Sep 2010, 13:10
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SnehaC wrote: The first person on the committee can be anyone of the 8.
The second person on the committee can be only one out of 6 (the first person with her or his sibling excluded).
The third person can be selected only out of 4 (first two members and their siblings excluded), giving 8*6*4 permutations. These are called permutations because we count every distinct combination of 3 people 6 times (6=3!). For example, we count {Ann, Bill, and Jane} as {Ann, Bill, Jane}, {Ann, Jane, Bill}, {Jane, Ann, Bill}, {Jane, Bill, Ann}, {Bill, Ann, Jane}, {Bill, Jane, Ann}. To get the number of combinations, divide the number of permutations by 6: 8*6*4 / (6) = 32 combinations.
I understand uptil the last part but I don't understand why we're diving by 6? Can anyone clarify this for me? We need to divide 8*6*4=192 by the factorial of the # of people - 3! to get rid of duplications 8*6*4=192 contains ---> \frac{192}{3!}=32 - correct answer. Consider example with smaller number: 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 as proposed in the solution you posted: The first person on the committee can be anyone of the 4. The second person on the committee can be only one out of 2 (the first person with her or his sibling excluded). So we'll get: 4*2=8, so more than 4, which means that 8 contains some duplications. And to get the right answer we should divide 8 by 2! --> 8/2!=4. It's because if you pick A1 for the first pick and than pick B1 for the second you'll get the committee {A1, B1} but if you'll pick B1 for the first pick and then A1 you'll get the exact same committee {A1, B1} (dividing by the factorial of the # of people in committee you'll exclude this double countings). Hope it helps.
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Re: If there are four distinct pairs of brothers and sisters [#permalink]
06 Sep 2010, 18:54
Bunuel, i like the way u use quick formulas for permutations and combinations..can u let me know how can i get these formulas? i want to strengthen my skills on this subject..
Thanks!
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Re: If there are four distinct pairs of brothers and sisters [#permalink]
07 Sep 2010, 00:04
Good explaination
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Re: If there are four distinct pairs of brothers and sisters [#permalink]
07 Sep 2010, 04:11
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Re: If there are four distinct pairs of brothers and sisters [#permalink]
07 Sep 2010, 05:46
Bunnuel - Thanks so much! The way you explained it was crystal clear  +1!
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Re: If there are four distinct pairs of brothers and sisters [#permalink]
23 Jan 2011, 14:04
Bunuel ....thanks a tonne.
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Re: If there are four distinct pairs of brothers and sisters [#permalink]
27 Dec 2012, 19:38
How many ways to select 3 of the pairs with representative in the group from 4 pairs? 4!/3!1! = 4
How many ways to select a representative from each pair? 2 x 2 x 2 = 8
4*8 = 32
Answer: C
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There are four distinct pairs of brothers and sisters. [#permalink]
25 Jan 2013, 01:05
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 I got 32 with 4c3 * 2*2*2. But why is 8c1 * 6c1 * 4c1 wrong ??/ Someone please explain ...
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Re: There are four distinct pairs of brothers and sisters. [#permalink]
25 Jan 2013, 01:26
shikhar 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
I got 32 with 4c3 * 2*2*2.
But why is 8c1 * 6c1 * 4c1 wrong ??/
Someone please explain ... You need to divide C^8_1*C^6_1*C^4_1with 3! to eliminate the duplicates (as the order in the arrangement does not matter). Number of ways a committee of 3 be formed and NOT have siblings in it = \frac{C^8_1*C^6_1*C^4_1}{3!} = \frac{8*6*4}{6}=32Hence choice(C) is the answer.
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Re: There are four distinct pairs of brothers and sisters. [#permalink]
25 Jan 2013, 05:01
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Re: If there are four distinct pairs of brothers and sisters [#permalink]
25 Jan 2013, 07:41
Total number of possible committees = 56
Total number of possible committees with a sibling pair = 6C1 x 4 = 24
Therefore, total # of committees w/out a sibling pair = 56-24 = 32
Answer: C
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Re: If there are four distinct pairs of brothers and sisters
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25 Jan 2013, 07:41
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