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A tea party is arranged for 16 people along the two sides of

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A tea party is arranged for 16 people along the two sides of [#permalink] New post 02 May 2005, 05:37
A tea party is arranged for 16 people along the two sides of a long table with 8 chairs on each side. Four men wish to sit on one particular side and 2 on the other side. In how many ways can they be seated?
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Re: permutations - good one [#permalink] New post 02 May 2005, 12:09
Economiser wrote:
A tea party is arranged for 16 people along the two sides of a long table with 8 chairs on each side. Four men wish to sit on one particular side and 2 on the other side. In how many ways can they be seated?


We can tackle this one too.

Firts, we select 4 chairs on one side ordered - 8!/4!
Second, we select 2 chairs on the other side ordered - 8!/6!
We are left with 10 chairs, ordering them gives - 10!

So it is (8!/4!)*(8!/6!)*10!
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Re: permutations - good one [#permalink] New post 02 May 2005, 22:36
= (10C4) (8!) (10C6)(8!)
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 [#permalink] New post 03 May 2005, 04:37
I have a slightly different version :

8C4.10C4.8C2.6!
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 [#permalink] New post 04 May 2005, 05:32
twixt wrote:
I have a slightly different version :

8C4.10C4.8C2.6!


My soln (twixt , I hope I haven't messed this one up).

The 4 ppl on side choose 4 out of 8 seats , so do the other 2 ppl.
The 4 and 2 then permute amongst themselves.
The remaining occupy the remaining 10 seats and then permute among themselves.

So, my ans.
8C4.8C2.4!.2!.10!

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 [#permalink] New post 05 May 2005, 11:21
I got 8C4 * 8C2 * 10!
What is the OA?
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 [#permalink] New post 05 May 2005, 12:44
my answer:8C4*4!*8C2*2!*10!
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 [#permalink] New post 05 May 2005, 14:48
Guys I am totally away from your solutions

2*[7!*2!*5!*4!]
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 [#permalink] New post 06 May 2005, 01:01
I get 2*(10 C 6)* 8! * 8!
What's the OA? :roll:
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Slightly complicated.. [#permalink] New post 06 May 2005, 19:02
The problem involves two parts.

part 1 - Combinations:

4 want to sit on one paticular side - so this can only be done in one way.
2 want to sit on another side - this can be done in only one way.

That leaves 10. These 10 (first worry about breaking them into one side or the other. We will worry about seating - arrangements in Part 2 of the solution) can be divided into two groups (group of 4 and group of 6).

Out of 10, 4 need to go on one side (and the rest automatically go on the other side). So, this alone can be done in 10C4 (or if you work on the other side, you will end up with 10C6. Note that 10C4 and 10C6 are the same). So, part one of the answer is 10C4 (or 10C6 both are equal).

Now, the problem asks about howmany ways they can be seated.

Part 2 - Permutations:

As far as arranging them, the problem did not say the team of 4 has to sit together (the problem only states that the team of 4 want to sit on a particular side - together or not).

therefore, one side of the table can be seated in 8! ways and with the same logic the other side can be seated in 8! ways.

So, the answer should be 10C4 * 8! * 8!.

If the problem stated that the team of 4 should be seated on one particular side together and team of 2 shoudl be seated on another side together, then
the answer might be

10C4 * 5! * 7! * 4! * 2!.

if the problem stated that the team of 4 should be seated together on one side and team of 2 should be seated together on the other side (without saying a 'particular' side), then anwer is

10C4 * 5! * 7! * 4! * 2! * 2!.

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 [#permalink] New post 06 May 2005, 19:46
It is 10C4*8!*8!


Here is my solution

Let 4 persons sit on one side(I) and 2 on other side(II).

Remaining 10 are there to be seated out of which 4
have to be selected to be seated on (I) ==> 10C4 ways

rest of the six have to sit on other side in 6C6 ways=1

Again on either sides the can be arranging themselves in 8! ways each.

REMEMBER: Here the question never asked for the 4 persons and the 2 persons sitting together. Rather it said that they are to be on the same end.

So 10C4*8!*8!
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Agree with you. [#permalink] New post 07 May 2005, 15:02
See my earlier post.
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Re: Agree with you. [#permalink] New post 07 May 2005, 16:00
Srinivas wrote:
See my earlier post.


Srinivas: Yes you were right....
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Re: permutations - good one [#permalink] New post 10 May 2005, 13:15
Tyr wrote:
Economiser wrote:
A tea party is arranged for 16 people along the two sides of a long table with 8 chairs on each side. Four men wish to sit on one particular side and 2 on the other side. In how many ways can they be seated?


We can tackle this one too.

Firts, we select 4 chairs on one side ordered - 8!/4!
Second, we select 2 chairs on the other side ordered - 8!/6!
We are left with 10 chairs, ordering them gives - 10!

So it is (8!/4!)*(8!/6!)*10!


Srinivas - I guess TYR reply is right. How come this a combination problem this is fully a permutation problem. :!:

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Re: permutations - good one [#permalink] New post 11 May 2005, 15:49
Rearrange 8C4 * 8C6 * 10! to see that it is indeed 8! * 8! * 10C6

-Srinivas


Jacques wrote:
Tyr wrote:
Economiser wrote:
A tea party is arranged for 16 people along the two sides of a long table with 8 chairs on each side. Four men wish to sit on one particular side and 2 on the other side. In how many ways can they be seated?


We can tackle this one too.

Firts, we select 4 chairs on one side ordered - 8!/4!
Second, we select 2 chairs on the other side ordered - 8!/6!
We are left with 10 chairs, ordering them gives - 10!

So it is (8!/4!)*(8!/6!)*10!


Srinivas - I guess TYR reply is right. How come this a combination problem this is fully a permutation problem. :!:
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Re: permutations - good one [#permalink] New post 11 May 2005, 18:25
My apologies, read this as "rearrange (8!/4!)*(8!/6!)*10 to see that it is indeed 8! * 8! * 10C6"

(
Srinivas wrote:
Rearrange 8C4 * 8C6 * 10! to see that it is indeed 8! * 8! * 10C6

-Srinivas


Jacques wrote:
Tyr wrote:
Economiser wrote:
A tea party is arranged for 16 people along the two sides of a long table with 8 chairs on each side. Four men wish to sit on one particular side and 2 on the other side. In how many ways can they be seated?


We can tackle this one too.

Firts, we select 4 chairs on one side ordered - 8!/4!
Second, we select 2 chairs on the other side ordered - 8!/6!
We are left with 10 chairs, ordering them gives - 10!

So it is (8!/4!)*(8!/6!)*10!


Srinivas - I guess TYR reply is right. How come this a combination problem this is fully a permutation problem. :!:
Re: permutations - good one   [#permalink] 11 May 2005, 18:25
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