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How many unique arrangements can be made of 3 people who

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Manager
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How many unique arrangements can be made of 3 people who [#permalink]

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04 Dec 2003, 11:51
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

How many unique arrangements can be made of 3 people who have to sit in 6 chairs? (They can have empty chairs between them)

Is there a general permutation formula for eliminating the double counting that will happen because of the spaces?

Thanks.
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04 Dec 2003, 12:11
bluefox420 wrote:
How many unique arrangements can be made of 3 people who have to sit in 6 chairs? (They can have empty chairs between them)

Is there a general permutation formula for eliminating the double counting that will happen because of the spaces?

Thanks.

i think it would be just 6p3 = 6! /3! = 6*5*4= 120 ways

i dont think there is a double counting here..

each position is unique.

thanks
praetorian
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04 Dec 2003, 12:19
"each position is unique."

But not each space!
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04 Dec 2003, 12:40
6!/(6-3)! is the formula we would use if there were no blanks or spaces.

But that just gives us the order of the people, and doesn't account for spaces between them. People could be arranged:
ABCXXX XABCXX XXABCX XXXABC
ABXCXX XABXCX XXABXC
ABXXCX XABXXC
ABXXXC

AXBCXX XAXBCX XXAXBC
AXBXCX XAXBXC
AXBXXC

AXXBCX XAXXBC
AXXBXC

AXXXBC

20 different ways

The solution is 20*6!/(6-3)!

But I don't know the formula, so I can't answer the question.
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04 Dec 2003, 13:35
Can we think of this problem as similar to finding the possible anagrams of the word ABCXXX?

If we do this, does this become 6!/3! ?

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04 Dec 2003, 14:53
Can we think of this problem as similar to finding the possible anagrams of the word ABCXXX?
A-ha!

Yes and no. If you did that, you would have all possible permutations of ABC and the three spaces, but what about C, D, and E?

Maybe do this in two steps:
1. Number of COMBINATIONS from 6c3, which is 20; multiply by
2. Number of anagrams from ABCXXX, which is 6!/(3!*1!*1!*1!), which is 120.

Total=2400.

Math is the same as my previous solution but this seems more sensible.

And the real killer is that a reasonably dedicated high school senior could this in his sleep...
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04 Dec 2003, 15:09
stoolfi wrote:
Can we think of this problem as similar to finding the possible anagrams of the word ABCXXX?
A-ha!

Yes and no. If you did that, you would have all possible permutations of ABC and the three spaces, but what about C, D, and E?

Maybe do this in two steps:
1. Number of COMBINATIONS from 6c3, which is 20; multiply by
2. Number of anagrams from ABCXXX, which is 6!/(3!*1!*1!*1!), which is 120.

Total=2400.

Math is the same as my previous solution but this seems more sensible.

And the real killer is that a reasonably dedicated high school senior could this in his sleep...

hey stoolfi,

we can pick 3 chairs out of 6 in 20 ways...

and then *each* of these "Ways"...seat these three..

So these three can be seated in 3! ways..isnt it.

And we have 20 of these combinations.

so...the total for 20 combination is 20 * 3! = 120 ways

agreed?

thanks
praetorian
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04 Dec 2003, 15:30
we can pick 3 chairs out of 6 in 20 ways...

and then *each* of these "Ways"...seat these three..

So these three can be seated in 3! ways..isnt it.

And we have 20 of these combinations.

so...the total for 20 combination is 20 * 3! = 120 ways

If we had only three people, I would agree with you. But we have six people.
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04 Dec 2003, 15:43
stoolfi wrote:
we can pick 3 chairs out of 6 in 20 ways...

and then *each* of these "Ways"...seat these three..

So these three can be seated in 3! ways..isnt it.

And we have 20 of these combinations.

so...the total for 20 combination is 20 * 3! = 120 ways

If we had only three people, I would agree with you. But we have six people.

are we sure we have six people?
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04 Dec 2003, 16:27
Sorry if I worded the question ambiguously - it's 3 people in 6 chairs (we have 3 people...)
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04 Dec 2003, 16:34
bluefox420 wrote:
Sorry if I worded the question ambiguously - it's 3 people in 6 chairs (we have 3 people...)

there is no ambiguity..our friend stoolfi just missed that part.
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04 Dec 2003, 17:19
there is no ambiguity..our friend stoolfi just missed that part.

Indeed, he did.

That stoolfi guy is a moron who can't read!!!

Well, time to start a new thread.
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04 Dec 2003, 17:29
stoolfi wrote:
there is no ambiguity..our friend stoolfi just missed that part.

Indeed, he did.

That stoolfi guy is a moron who can't read!!!

Well, time to start a new thread.

Dont be so hard on yourself
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04 Dec 2003, 19:17
bluefox420 wrote:
Can we think of this problem as similar to finding the possible anagrams of the word ABCXXX?

If we do this, does this become 6!/3! ?

I agree with you. That's exactly what I did: P6/3!=6!/3!

Martin
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05 Dec 2003, 01:45
so hot a discussion, but the solution is simple. it is arrangements (similar to combinations but order is important).

Out of 6, we have to take 3 places+order is important
6A3=6!/3!=4*5*6=120

or

out of 6, we take three places without order, and sit 3 persons in 3! ways

6C3*3!=20*6=120
05 Dec 2003, 01:45
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