If there are four distinct pairs of brothers and sisters, then in how

Question

From four sibling pairs, each consisting of a brother and a sister, how many three-member committees can be formed without including siblings from the same pair?

A. 8
B. 24
C. 32
D. 56
E. 80

M02-05

Bunuel · Accepted Answer

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.

Bunuel · Answer

Official Solution:

From four sibling pairs, each consisting of a brother and a sister, how many three-member committees can be formed without including siblings from the same pair?

A. 8
B. 24
C. 32
D. 56
E. 80

Since the committee must not include siblings, only one "representative" from each sibling pair can be chosen. The number of ways to select three pairs of siblings that will send one representative to the committee is represented by  C^3_4  (choosing 3 pairs from 4, which will be granted the right to send a representative).

For each of these three pairs, there are two options for the representative: either the brother or the sister. This results in  2 * 2 * 2 = 2^3  possibilities.

Therefore, the total number of ways to form the committee is  C^3_4*2^3=32 .

Answer: C

muralimba · Answer

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

SnehaC · Answer

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?

vdixit · Answer

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!

Bunuel · Answer

Probability and Combinatorics chapters of Math Book: math-probability-87244.html math-combinatorics-87345.html You can also see Probability and Combinatorics questions to practice at: viewforumtags.php Hope it helps.

GMATmission · Answer

Bunnel, how you came to the conclusion that 8C1x6C1x4C1 should be divided by 3! to get rid of duplications. Why not 4! or 2!. Please explain.

Bunuel · Answer

We divide by the number of members in the committee, so by 3!. Consider this: 8C1*6C1*4C1 will give you all committees of ABC possible - (ABC), (ACB), (BAC), (BCA), (CAB) and (CBS) which are the SAME committee of 3 (3 distinct letters can be arranged in 3! ways). So we should divided 8C1*6C1*4C1 by 3!. This question and the same doubt you have is also discussed here: https://gmatclub.com/forum/if-there-are- ... 99992.html Similar questions to practice: https://gmatclub.com/forum/ps-combinations-94068.html https://gmatclub.com/forum/ps-combinations-101784.html https://gmatclub.com/forum/committee-of-88772.html https://gmatclub.com/forum/if-4-people-a ... 99055.html https://gmatclub.com/forum/combination-p ... 98533.html Hope it helps.

mbaiseasy · Answer

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

PrashantPonde · Answer

You need to divide  C^8_1*C^6_1*C^4_1 with  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}=32

Hence choice(C) is the answer.

him1985 · Answer

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

No. of ways to select 3 people = 3!

so 192/3! = 32

GMATinsight · Answer

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

paidlukkha · Answer

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?

GMATinsight · Answer

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!!!

GMATinsight · Answer

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

Total favourable cases = 8C3 - (4*6) = 56 - 24 = 32

GyMrAT · Answer

# of ways to from a committee of 3 with no siblings on it = (8 * 6 * 4)/3! = 32

The First person in the committee can be chosen in 8 ways.
The Second person in the committee can be chosen in 6 ways, since we will remove the sibling of first person selected & then choose.
The Third person in the committee can be chosen in 4 ways, since we will remove the sibling of second person selected, as well & then choose.

Now we are asked for the # of ways to form a committee & not the arrangement of the persons, hence order does not matter. To account for this we divide by 3!, which is the # of possible arrangements of 3 persons.

Answer C.

Thanks,
GyM

Bixy34 · Answer

Hi, I am solving it this way and unable to understand what I am doing wrong- 
Choosing 1 sibling pair out of 4- 4C1
Choosing 1 person out of the sibling pair- 2C1
Choosing the remaining 2 committee members out of the remaining 6 people- 6C2

i.e 4C1 x 2C1 x 6C2 = 120 ways.

Can someone help me understand what's wrong with my approach?

Krunaal · Answer

The remaining 2 committee members you choose out of 6 could include a sibling pair. Also, selecting one sibling pair and then one person doesn't help, the correct way to approach this problem is this: https://gmatclub.com/forum/if-there-are ... ml#p770967 Alternatively, You can also approach by getting all possible cases and then subtracting the one's that are not valid, All possible cases => 8C3 = 56 Invalid cases = Selecting a sibling pair: 4C1, and the third person out of remaining 6: 6C1 => 4*6 = 24 56 - 24 = 32

If there are four distinct pairs of brothers and sisters, then in how

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