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by satishchandra » Thu Nov 03, 2011 9:12 pm
nsamkari wrote:IN HOW MANY WAYS COULD I ARRANGE 3 BOYS AND 4 GIRLS AROUND A CIRCULR TABLE?
If there are no constraints then we can just treat 3 Boys and 4 girls as 7 people
6! ways

Always Remember: (i)To arrange 'n' items in a row: n! ways
(ii) To arrange 'n' items around a circle: (n-1)! ways
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by nsamkari » Thu Nov 03, 2011 9:18 pm
THANKS

WHAT'S THE ANSWER IF I WANT TO START WITH GIRL?
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by nsamkari » Thu Nov 03, 2011 9:32 pm
ALSO WHAT IS THE ANSWER IF I WANT TO ARRANGE IT IN ORDER OF G,B,,G....
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by satishchandra » Thu Nov 03, 2011 9:41 pm
nsamkari wrote: WHAT'S THE ANSWER IF I WANT TO START WITH GIRL?
You cant start a circular row with a girl. Just imagine a group of people sitting in a circle. If I ask you who's sitting in the 1st place? Can you take anyone's name?
However, If you are arranging them in a row, you can have a girl in first place.

The correct form of question can be
If I send 1 girl in circular table to sit 1st. What are the number of ways for the rest of people to occupy seats?
I guess it would still be 6!
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by shankar.ashwin » Thu Nov 03, 2011 9:41 pm
You cannot start with a particular person in circular arrangements, imagine 7 people sitting in a circular table, how could you say which is the first position? Unless each chair is numbered or something like that, we can't tell.
nsamkari wrote:THANKS

WHAT'S THE ANSWER IF I WANT TO START WITH GIRL?
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by satishchandra » Thu Nov 03, 2011 10:08 pm
nsamkari wrote:ALSO WHAT IS THE ANSWER IF I WANT TO ARRANGE IT IN ORDER OF G,B,,G....
This is a good GMAT like question, which makes sense.

Now see.
You have 4 Girls and 3 Boys.
Couple 1 girl and 1 boy and form a Group. you will have 3 groups + 1 girl
There are now 4 elements.
We can arrange 4 elements in circle in 3! ways. Its not over.
In a group, 1 boy+1 Girl can be arranged in 2! ways.
In, total = 3!2! = 12 ways
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by anuu » Fri Nov 04, 2011 7:18 am
satishchandra wrote:
nsamkari wrote:ALSO WHAT IS THE ANSWER IF I WANT TO ARRANGE IT IN ORDER OF G,B,,G....
This is a good GMAT like question, which makes sense.

Now see.
You have 4 Girls and 3 Boys.
Couple 1 girl and 1 boy and form a Group. you will have 3 groups + 1 girl
There are now 4 elements.
We can arrange 4 elements in circle in 3! ways. Its not over.
In a group, 1 boy+1 Girl can be arranged in 2! ways.
In, total = 3!2! = 12 ways


Hi,

I've a doubt in the following point..Can you pls clarify it?

In a group 1 boy +1 girl can be arranged in 2! ways..since there are 3 groups,
there would be 3*2! ways (6 ways)to arrange the boys&girls in the 3 groups.

Total ways = 3!*6 = 36 ways

Thank you,
anu
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by GMATGuruNY » Fri Nov 04, 2011 9:13 am
nsamkari wrote:ALSO WHAT IS THE ANSWER IF I WANT TO ARRANGE IT IN ORDER OF G,B,,G....
If there are 4 girls and 3 boys, there is no way to arrange them in a circle so that the genders alternate.
If we start with a boy, we get:
BGBGBGG, in which case the last 2 girls sit in adjacent seats.
If we start with a girl, we get:
GBGBGBG, in which case the first girl and the last girl sit in adacent seats.

To alternate between boys and girls, we must have the same of number of boys as we have girls.
Given 3 girls and 3 boys:
Number of ways to arrange the 3 girls in a circle = (3-1)! = 2.
Number of ways to arrange the 3 boys in the remaining 3 seats = 3! = 6.
To combine the options above, we multiply:
2*6 = 12.

Another approach:
In a circle, it doesn't matter where the first child sits, since there is no first seat or last seat.
What matters is how everyone else sits RELATIVE to the first child.
Thus, we can place one of the boys at the table and count the number of ways to arrange the remaining children RELATIVE to the first boy.

Moving clockwise around the table:
Number of options for the next seat = 3. (Must be one of the 3 girls.)
Number of options for the next seat = 2. (Must be one of the 2 remaining boys.)
Number of options for the next seat = 2. (Must be one of the 2 remaining girls.)
Number of options for the next seat = 1. (Must be the one remaining boy.)
Number of options for the next seat = 1. (Must be the one remaining girl.)
To combine the options above, we multiply:
3*2*2*1*1 = 12.
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