Question: $12$ guests at a dinner party are to be seated along a circular table. Suppose that the master and mistress of the house have fixed seats opposite to one another. There are two specified guests who must always be placed next to one another. What is the number of ways in which the company can be placed?

My Method:
Let us first seat the $12$ guests. $2$ guests are always beside each other hence considering $11$ guests, the no. of combinations possible are $11!$. But the $2$ people can be arranged in $2$ ways, so total no. of combinations is given by $2*11!$.
Now the master can be seated between any $2$ people except the pair to be seated together. Hence, the no. of possibilities to seat him becomes $11$. The mistress is always opposite to him hence she would not contribute to the no. of total ways.

This gives,
The total no. of ways $=2*11*11!$

The Answer given:
First we seat the $2$ specified people in $2*10$ ways and the remaining $10$ people can be arranged in $10!$ ways. So total no. of ways $=2*10*10!$

But I don't understand what does the answer state and how my way is wrong. Can anybody help me understand this...

  • $\begingroup$ Not sure I get the set up. How many people are sitting at this table? $12$ or $14$? $\endgroup$ – lulu Oct 18 '16 at 15:10
  • $\begingroup$ @lulu Well I guess 14 $\endgroup$ – Osheen Sachdev Oct 18 '16 at 15:11
  • $\begingroup$ @lulu 12 guests and the two heads of the house. That makes 14. If 12 people in total, only $2×8$ ways the special pair can be seated. $\endgroup$ – Parcly Taxel Oct 18 '16 at 15:11
  • $\begingroup$ I think you (OP) misread the question to mean that the hosts could sit in any pair of opposite seats. It's a plausible reading, but the other reading, which the answer indicates is the intended one, is that the two hosts are already assigned their seats; i.e., the only question is how many different ways the guests can be assigned their seats. $\endgroup$ – Brian Tung Oct 18 '16 at 16:42
  • $\begingroup$ @BrianTung: How does it affect, if the seats are taken to be unnumbered, as is customary in the absence of any information to the contrary ? $\endgroup$ – true blue anil Oct 18 '16 at 16:54

The book answer is correct

There is the master, mistress and $12$ guests, and in the absence of information to the contrary, it is customary to take chairs to be unnumbered, so there is only one way to seat the hosts opposite to each other.

In effect, the remaining $12$ seats become numbered with respect to the mistress at the $12$ o'clock position, say, and the master at $6$ o'clock position.

In between there are $6$ seats in each half, but the two specified people can only sit together in $2\cdot5$ ways in each half, i.e. $2\cdot10$ ways altogether.

The remaining $10$ guests can then be arranged in $10!$ ways

Putting everything together, ans $= 2*10*10!$

  • $\begingroup$ Sir master and mistress have two potion 12o clock and 6 o clock and vice versa $\endgroup$ – yuvraj singh Sep 22 at 8:09

The mistress is always opposite the master, therefore not only can the master not sit between the two specified guests, but he cannot sit between the pair opposite them, either. Also, before someone sits down, the round table has no "first" or "last" position, so each of your $11!$ options has $11$ identical rotations. In general, the number of ways to arrange $n$ objects in a circle is $(n-1)!,$ not $n!,$ because all positions in which the first object can be placed are indistinguishable.

To make this clearer, consider the simple case of $3$ objects. There are of course $3!=6$ ways of arranging them in a row: $$1.abc$$$$2.acb$$$$3.bac$$$$4.bca$$$$5.cab$$$$6.cba$$ However, when we put the objects in a circle, options $1,4,$ and $5$ are the same, as are $2,3,$ and $6$, so there are only $2!=2$ different circular arrangements.

  • $\begingroup$ Then again 2*10*11! and not 2*10*10! $\endgroup$ – Osheen Sachdev Oct 18 '16 at 15:08
  • $\begingroup$ Good point. Edited answer to address the other issue as well. $\endgroup$ – Gabriel Burns Oct 18 '16 at 15:10
  • $\begingroup$ If you seat the two particular people first, the computation will become more complicated, but must still yield 2*10*10! $\endgroup$ – true blue anil Oct 18 '16 at 17:32
  • $\begingroup$ @trueblueanil and it does. His comment was to a previous version of this answer. Read before you downvote. $\endgroup$ – Gabriel Burns Oct 18 '16 at 17:34
  • 1
    $\begingroup$ As to "what does the answer state", I started to address that (although it's not entirely clear what that part of the question means) in another edit. While I was working on it you posted your answer, which explained the correct derivation basically how I would have. I figured there was no reason for me to repeat what you had done, so I left it as is. Between the two of us, he got what he wanted: the correct approach (from you, and assuming that's what he meant by "what does the answer state"), and the problem with his (from me). $\endgroup$ – Gabriel Burns Oct 18 '16 at 17:55

The question is correct there is no mistake in the question.

My Answer is:

First we seat the master and the mistress in $2$ selected Seat by $1!$ Because the seat is already fixed for both the person and opposite to each other and thus these two seats divide the whole circle into two equal part $6$ on each side. Now seat the two guest by selecting $2$ consecutive seat the way is $(5+5)$ and arrange by $2!$ and the remaining $10$ person will sit on remaining $10$ seats by $10!$ So the final answer is $$(5+5)\cdot 2!\cdot 10!$$


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