P & C

[rahul.s]

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

I know it seems like a simple problem, but I opted for E, and the OA is C.

[thephoenix]

number of ways of selection of 3 out of 8 is 8 C 3=56

there are 4 pairs out of which one pair can be selected in 4C1=4 WAYS
the remaining third person has to be selcted from 6 left =6C1=6
tot=6*4=24

reqd ways is 56-24=32

[ajith]

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

I know it seems like a simple problem, but I opted for E, and the OA is C.

The First person can be chosen in 8 ways, the second person in 6 ways (the first persons sibling cannot be selected) and the third person in 4 ways

Total no of ways = 8*6*4 = 192. These include all the arrangement but we need only combination. So, to get the no of combination we have divide the arrangements by 3!

The committee can be formed in = 192/6 = 32 ways

[ajith]

https://www.beatthegmat.com/quant-questi ... tml#216008

[rahul.s]

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

I know it seems like a simple problem, but I opted for E, and the OA is C.

The First person can be chosen in 8 ways, the second person in 6 ways (the first persons sibling cannot be selected) and the third person in 4 ways

Total no of ways = 8*6*4 = 192. These include all the arrangement but we need only combination. So, to get the no of combination we have divide the arrangements by 3!

The committee can be formed in = 192/6 = 32 ways

i got the part up to 192. but why do we divide it by 3!?

[ajith]

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

I know it seems like a simple problem, but I opted for E, and the OA is C.

The First person can be chosen in 8 ways, the second person in 6 ways (the first persons sibling cannot be selected) and the third person in 4 ways

Total no of ways = 8*6*4 = 192. These include all the arrangement but we need only combination. So, to get the no of combination we have divide the arrangements by 3!

The committee can be formed in = 192/6 = 32 ways

i got the part up to 192. but why do we divide it by 3!?

Discussed in detail in the link provided above, please have a look and if you still have a doubt revert

[GMATinsight]

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

I know it seems like a simple problem, but I opted for E, and the OA is C.

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

[Brent@GMATPrepNow]

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

A brother and sister form a FAMILY.
So, we need a SINGLE representative from 3 of the 4 families.

Take the task of creating the committee, and break it into stages.

Stage 1: Select 3 families (from which we will choose representatives)
Since the order in which we select the families does not matter, we can use combinations.
We can select 3 families from 4 families in 4C3 ways (4 ways)
So, we can complete stage 1 in 4 ways

If anyone is interested, we have a free video on calculating combinations (like 4C3) in your head: https://www.gmatprepnow.com/module/gmat-counting?id=789

Stage 2: From one family, select a representative
There are 2 siblings to choose from, so we can complete this stage in 2 ways.

Stage 3: From another family, select a representative
There are 2 siblings to choose from, so we can complete this stage in 2 ways.

Stage 4: From the last remaining family, select a representative
There are 2 siblings to choose from, so we can complete this stage in 2 ways.

By the Fundamental Counting Principle (FCP), we can complete all 4 stages (and thus create a committee) in (4)(2)(2)(2) ways ([spoiler]= 32 ways[/spoiler])

Answer: C
--------------------------

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. For more information about the FCP, watch our free video: https://www.gmatprepnow.com/module/gmat-counting?id=775

Then you can try solving the following questions:

EASY
- https://www.beatthegmat.com/what-should- ... 67256.html
- https://www.beatthegmat.com/counting-pro ... 44302.html
- https://www.beatthegmat.com/picking-a-5- ... 73110.html
- https://www.beatthegmat.com/permutation- ... 57412.html
- https://www.beatthegmat.com/simple-one-t270061.html
- https://www.beatthegmat.com/mouse-pellets-t274303.html


MEDIUM
- https://www.beatthegmat.com/combinatoric ... 73194.html
- https://www.beatthegmat.com/arabian-hors ... 50703.html
- https://www.beatthegmat.com/sub-sets-pro ... 73337.html
- https://www.beatthegmat.com/combinatoric ... 73180.html
- https://www.beatthegmat.com/digits-numbers-t270127.html
- https://www.beatthegmat.com/doubt-on-sep ... 71047.html
- https://www.beatthegmat.com/combinatoric ... 67079.html


DIFFICULT
- https://www.beatthegmat.com/wonderful-p- ... 71001.html
- https://www.beatthegmat.com/ps-counting-t273659.html
- https://www.beatthegmat.com/permutation- ... 73915.html
- https://www.beatthegmat.com/please-solve ... 71499.html
- https://www.beatthegmat.com/no-two-ladie ... 75661.html
- https://www.beatthegmat.com/laniera-s-co ... 15764.html

Cheers,
Brent