Dinner Party [Repost for more discussion]

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dxgamez: Senior | Next Rank: 100 Posts; Posts: 65; Joined: Wed Nov 25, 2009 6:33 pm; Thanked: 3 times

Dinner Party [Repost for more discussion]

by dxgamez » Wed Apr 14, 2010 5:51 pm

At a dinner party, 5 people are to be seated around a circular table. Two seating arrangements are said to be different only when the positions of people are different relative to each other. What is the total number of different possible seating arrangements for the group?

A) 5
(B) 10
(C) 24
(D) 32
(E) 120

** I don't understand the answer given. By the way, I'm re-posting(this qn is already in the forum) so as to get more discussions on the permutations. Cant seem to grasp the idea..

Thanks.

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Source: Beat The GMAT — Quantitative Reasoning | Wed Apr 14, 2010 5:51 pm

4GMAT_Mumbai: Master | Next Rank: 500 Posts; Posts: 161; Joined: Mon Apr 05, 2010 9:06 am; Location: Mumbai; Thanked: 37 times

by 4GMAT_Mumbai » Wed Apr 14, 2010 7:15 pm

Hi dxgames, I shall try my best.

Arranging n people in a circle is equivalent to arranging (n-1) people in a straight row. Arranging (n-1) people in a straight row can be done in (n-1)! ways. The last person has only one seat left - in other words, she has no choice left. She sits in the last chair left out (only 1 way of doing this). Hence, the total works out to (n-1)! times 1 = (n-1)!

This assumes that clockwise and anticlockwise seating patterns of a particular sequence of people are counted as two unique arrangements. If they are not counted as unique, then the number of ways boils gets halved to (n-1)! / 2.

The question states that 'Two seating arrangements are said to be different only when the positions of people are different relative to each other'. This seems to suggest that clockwise and anticlockwise arrangements are to be counted as two arrangements. i.e., having a person on my left is different from having the same person on my right.

Hence, IMO the answer is [spoiler](n-1)! and thus 24.[/spoiler]

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dxgamez: Senior | Next Rank: 100 Posts; Posts: 65; Joined: Wed Nov 25, 2009 6:33 pm; Thanked: 3 times

by dxgamez » Wed Apr 14, 2010 7:48 pm

Thanks 4GMAT, much clearer now.

Just got stuck into visualizing why arranging n people in a circle is equivalent to arranging n-1 ppl in a straight row.

I'm still visualizing the circle is linear like a straight row. Could you help to drill it down for me?

Seems permutations/combinations are kinda weak for me.

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4GMAT_Mumbai: Master | Next Rank: 500 Posts; Posts: 161; Joined: Mon Apr 05, 2010 9:06 am; Location: Mumbai; Thanked: 37 times

by 4GMAT_Mumbai » Wed Apr 14, 2010 8:41 pm

Anytime, dxgamez.

I have tried putting it in a visual attachment. There are 10 people in the 1st figure. Let the question be the number of ways in which they can sit in the circular table.

Where do I start? Let A sit put on the table. The other 9 people stand in a straight line in front of the table (as in 2nd part of the figure). I am left with 9 people labelled B - J.

The 9 people B - J can be arranged in 9! ways in a straight line. For each of these 9! ways, they retain their formation and go and sit around A. i.e., the guy on the left most position sits to the right of A. The guy 2nd from left sits in the second position to the right of A ... the guy on the right most position sits to the left of A.

Thus, the whole thing can be done in 9! ways. (n-1)!

Questions you could ask:

1. Would it not matter if B decides to stay put on the table instead of A.

It would not matter because it is a circular table. It has no starting point nor an ending point.

2. Would it not matter if A decides to stay put in any other chair and not the one she is currently sitting in. Would it not result in more arrrangements?

Not quite - because of the same reason. The circle has neither a starting point nor an ending point. In an particular arrangement, if everybody decides to move to the chair on their left hand side that does not mean that we have got a new arrangement.

Hope this helps. Feel free to pm me if any further help is needed. I shall try my best.

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