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Find the number of ways in which four men, two women and a

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Find the number of ways in which four men, two women and a [#permalink] New post 26 Dec 2010, 07:39
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Find the number of ways in which four men, two women and a child can sit at a table if the child is seated between two women.
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Re: Circular Permutation Problem [#permalink] New post 26 Dec 2010, 07:59
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subhashghosh wrote:
Hi

Could someone please help me with this, I am getting an answer 240. But I'm not sure if I'm correct.

Find the number of ways in which four men, two women and a child can sit at a table if the child is seated between two women.

Regards,
Subhash


Note:
The number of arrangements of n distinct objects in a row is given by n!.
The number of arrangements of n distinct objects in a circle is given by (n-1)!.

We have M, M, M, M, W, W, C --> glue two women and the child so that they become one unit and the child is between women: {WCW}. Now, these 5 units: {M}, {M}, {M}, {M}, {WCW} can be arranged around the table in (5-1)!=4! ways and the women within their unit can be arranged in 2 ways {W1, C, W2} or {W2, C, W1} so total # of arrangement is 4!*2=48.

Answer: 48.

240 would be the answer if the arrangement were in a row: {M}, {M}, {M}, {M}, {WCW} can be arranged in a row in 5! ways and the women within their unit can be arranged in 2 ways {W1, C, W2} or {W2, C, W1} so total # of arrangement is 5!*2=240.

P.S. Please read and follow: how-to-improve-the-forum-search-function-for-others-99451.html

So please provide answer choices for PS questions.
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Re: Circular Permutation Problem [#permalink] New post 26 Dec 2010, 08:23
Hi

Thanks for your reply. And I did not have the answer, so could not post it, apologies.

One question, if the places in table are numbered, wouldn't it become a case like arranging the members in the stated manner in a line, in which case the answer 240 is valid ?

Regards,
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Re: Circular Permutation Problem [#permalink] New post 26 Dec 2010, 08:48
Expert's post
subhashghosh wrote:
Hi

Thanks for your reply. And I did not have the answer, so could not post it, apologies.

One question, if the places in table are numbered, wouldn't it become a case like arranging the members in the stated manner in a line, in which case the answer 240 is valid ?

Regards,
Subhash


If the chairs are numbered and one specific arrangement and the same arrangement but shifted by one position are considered different then the answer will simply be 48*7.

That's because the difference between placement in a row and that in a circle is following: if we shift all object by one position, we will get different arrangement in a row but the same relative arrangement in a circle.
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Re: Circular Permutation Problem [#permalink] New post 26 Dec 2010, 09:06
Bunuel wrote:
subhashghosh wrote:

240 would be the answer if the arrangement were in a row: {M}, {M}, {M}, {M}, {WCW} can be arranged in a row in 5! ways and the women within their unit can be arranged in 2 ways {W1, C, W2} or {W2, C, W1} so total # of arrangement is 5!*2=240.


For this szenario I received 744 different cases, including the cases when the child and a man or men are between the 2 women.

{M}, {M}, {M},{WCMW} 4*3*2*2*2=96
{M}, {M},{WCMMW} 3*2*3*2=72
{M},{WCMMMW}2*4*3*2*2=96
{WCMMMMW}5*4*3*2*2=240
These possibilities together with Bunuels possibilities when only the child is btw the women {WCW} gives 504+240=744

Please correct me if I am wrong!
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Re: Circular Permutation Problem [#permalink] New post 26 Dec 2010, 09:15
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medanova wrote:
Bunuel wrote:
subhashghosh wrote:

240 would be the answer if the arrangement were in a row: {M}, {M}, {M}, {M}, {WCW} can be arranged in a row in 5! ways and the women within their unit can be arranged in 2 ways {W1, C, W2} or {W2, C, W1} so total # of arrangement is 5!*2=240.


For this szenario I received 744 different cases, including the cases when the child and a man or men are between the 2 women.

{M}, {M}, {M},{WCMW} 4*3*2*2*2=96
{M}, {M},{WCMMW} 3*2*3*2=72
{M},{WCMMMW}2*4*3*2*2=96
{WCMMMMW}5*4*3*2*2=240
These possibilities together with Bunuels possibilities when only the child is btw the women {WCW} gives 504+240=744

Please correct me if I am wrong!


I think the question means that ONLY the child must be between two women.
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Re: Circular Permutation Problem [#permalink] New post 26 Dec 2010, 09:22
Ok, I see.
If the question were stated for a row and not a table, and if there was the ONLY requirement, would that be correct?
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Re: Circular Permutation Problem [#permalink] New post 26 Dec 2010, 09:38
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medanova wrote:
Ok, I see.
If the question were stated for a row and not a table, and if there was the ONLY requirement, would that be correct?


If... If... This question is already not a GMAT type...

But anyway your solution is still wrong:

{M}, {M}, {M}, {M}, {WCW}: 5!*2=240;
{M}, {M}, {M}, {WCMW}: 4!*4C1*2!*2;
{M}, {M}, {WCMMW}: 3!*4C2*3!*2;
{M}, {WCMMMW}: 2!*4C3*4!*2;
{WCMMMMW}: 5!*2.
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Re: Circular Permutation Problem [#permalink] New post 20 Jan 2011, 08:38
Another great explanation. Thank you Bunuel

Bunuel wrote:
subhashghosh wrote:
Hi

Could someone please help me with this, I am getting an answer 240. But I'm not sure if I'm correct.

Find the number of ways in which four men, two women and a child can sit at a table if the child is seated between two women.

Regards,
Subhash


Note:
The number of arrangements of n distinct objects in a row is given by n!.
The number of arrangements of n distinct objects in a circle is given by (n-1)!.

We have M, M, M, M, W, W, C --> glue two women and the child so that they become one unit and the child is between women: {WCW}. Now, these 5 units: {M}, {M}, {M}, {M}, {WCW} can be arranged around the table in (5-1)!=4! ways and the women within their unit can be arranged in 2 ways {W1, C, W2} or {W2, C, W1} so total # of arrangement is 4!*2=48.

Answer: 48.

240 would be the answer if the arrangement were in a row: {M}, {M}, {M}, {M}, {WCW} can be arranged in a row in 5! ways and the women within their unit can be arranged in 2 ways {W1, C, W2} or {W2, C, W1} so total # of arrangement is 5!*2=240.

P.S. Please read and follow: how-to-improve-the-forum-search-function-for-others-99451.html

So please provide answer choices for PS questions.

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Re: Circular Permutation Problem [#permalink] New post 20 Jan 2011, 21:01
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Another way to look at the question.
(though let me point out first that the question doesn't specifically say 'circular table'. It just says 'sit at a table'. If it is a rectangular table, perhaps there are 4 chairs on one side, 3 on the other etc. Since you mentioned "Circular Permutation Problem" in the subject line, I am assuming it is meant to be a circular table.)

7 seats around a circular table, 7 people.
First I make the child sit anywhere in 1 way since all seats are the same. The two women can sit around him in 2! ways. Now 4 seats are left for 4 men and they can occupy them in 4! ways.
Total number of ways = 4!*2! = 48

Another thing, if the places are numbered, say 1, 2, 3 etc for the 7 seats, the number of arrangements will be 7*2!*4! = 336.
Make the child sit on any one of the 7 seats since all are unique now. The women sit around the child in 2! ways and the men sit on the rest of the 4 seats in 4! ways.
The reason why this number is greater than the number of arrangements in case of a row (240 ways) is because in a row, child cannot be in 1st or 7th position while in a circle, the child can sit on seat no 1 or seat no 7. So we have 2*2!*4! = 96 extra cases in case of numbered seats around a circular table.
Note: 240 + 96 = 336
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Re: Find the number of ways in which four men, two women and a [#permalink] New post 16 Aug 2014, 03:53
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Re: Find the number of ways in which four men, two women and a   [#permalink] 16 Aug 2014, 03:53
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