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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
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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) = 56-24 = 32

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New post 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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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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Is this a possible correct solution as well ?

8*6*4
-------
3!

Thanks you

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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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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: if-there-are-four-distinct-pairs-of-brothers-and-sisters-99992.html#p775925

Hope it helps.
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New post 07 Nov 2015, 04:49
[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.

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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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New post 12 May 2016, 07:53
4 pairs means 8 right?
Could this question be explained in a better way?

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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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New post 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 = 56-24= 32

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New post 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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New post 10 Jul 2017, 09:34
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
>> !!!

You do not have the required permissions to view the files attached to this post.

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New post 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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New post 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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New post 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/if-there-are- ... ml#p775925

Hope it helps.
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