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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: \(2*2*2=2^3\);

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:

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).

Re: If there are four distinct pairs of brothers and sisters [#permalink]

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06 Sep 2010, 12:29

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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?

Re: If there are four distinct pairs of brothers and sisters [#permalink]

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27 Dec 2012, 19:38

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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 \(4*8 = 32\)

AABBCCDD so ABC can come or BCD or CDA so 3! * 4 = 24 if A is not equal to A then it becomes 24*2 = 48 which ones am I missing?

Another method to solve this question is

Select any three out of these 8 individuals = 8C3 ways

Subtract the unwanted cases i.e. cases in which 2 of 3 selected have one sibling pair which can be selected as 4*6 4 = number of ways to select one sibling pair i.e. two individuals 6 = No. of ways of selecting one out of 6 remaining individuals to make a group of 3 alongwith 2 selected in previous step

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Re: If there are four distinct pairs of brothers and sisters [#permalink]

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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..

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!

Probability and Combinatorics chapters of Math Book:

Re: If there are four distinct pairs of brothers and sisters [#permalink]

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11 Jul 2014, 23:42

we can select 8 people for 1st place, 6 for second (only one from pair can be selected) and , 4 for 3rd So we can have total = 8*6*4 = 192 Now in above calculation, we have counted all no of ways. (ABC is different from ABE) so we have to divide the above value with no of ways we can select 3 people

Re: If there are four distinct pairs of brothers and sisters [#permalink]

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02 Nov 2014, 16:59

Bunuel wrote:

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:

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.

Hi Bunuel,

I have the same question as the other poster.

If we solve it as 8*6*4*2 == how are we creating duplicates? Aren't we eliminating the sibling by dropping down to 6 from 8 and so on?

You mention that we should divide by 2! in the above A1B1 solution. Does that mean that we would divide by 4! for the actual problem because there are four male and 4 female members or would we divide by 2 because of the sibling issue?

Re: If there are four distinct pairs of brothers and sisters [#permalink]

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02 Dec 2014, 07:46

Combinations : Choose unique 3 from 8 ( 4 pairs) rCn 8C3 = 56 ( Total number of combinations ) Condition : Non - Siblings and Siblings !!! Non = Total - Siblings But how to find the ways of siblings?? Can anyone explain it..
_________________

I welcome analysis on my posts and kudo +1 if helpful. It helps me to improve my craft.Thank you

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

LET ABCD are Boys and PQRS are their sisters respectively Case-1: All Boys – 4C3 = 4 Case-2: All Girl – 4C3 = 4 Case-3: 2 Boys and 1 girl – 4C2*2C1 = 12 Case-4: 2 Girl and 1 Boy – 4C2*2C1 = 12

Total Cases = 4+4+12+12 = 32

Answer: option C
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AABBCCDD so ABC can come or BCD or CDA so 3! * 4 = 24 if A is not equal to A then it becomes 24*2 = 48 which ones am I missing?

Point 1: You can't take them as AABBCCDD because in sibling couple also first individual and second individuals are treated differently Rather you should take them as A1A2 B1B2 C1C2 D1D2

In case you want to make 4 cases then All Ones i.e. three of A1, B1, C1, D1 which can happen in 4C3 ways All Twos i.e. three of A2, B2, C2, D2 which can happen in 4C3 ways Two ones and one Two i.e. 4C2*2C1 = 12 (2C1 is used to select one Two out of remaining twos who are not siblings of Ones selected Two Twos and one One i.e. 4C2*2C1 = 12 (2C1 is used to select one Two out of remaining twos who are not siblings of Ones selected

Total ways = 4+4+12+12 = 32

Point 2: you are using 3! in your solution which is completely redundant because there is no arrangement here. You only have to select 3 individuals out of 8 so the arrangement doesn't come in picture so use of 3! is completely incorrect on concept part

Point 3: I didn't understand why you used 4 in your solution.

I hope this helps!!!
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