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
P & C
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thephoenix
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ajith
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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 waysrahul.s 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?
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.
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
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rahul.s
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i got the part up to 192. but why do we divide it by 3!?ajith wrote: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 waysrahul.s 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?
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.
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
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ajith
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rahul.s wrote:i got the part up to 192. but why do we divide it by 3!?ajith wrote: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 waysrahul.s 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?
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.
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
Discussed in detail in the link provided above, please have a look and if you still have a doubt revert
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GMATinsight
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rahul.s 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?
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
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A brother and sister form a FAMILY.rahul.s 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?
A) 8
B) 24
C) 32
D) 56
E) 192
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
--------------------------
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