How many arrangements are possible around a table with ten guests

We assume that your table is a round table with ten chairs, one each
for each guest. Then the no. of arrangments is (10-1)!=9!=362880.

can you explain your answer in a way for someone without an extensive
math bachground would understand.yhank you

Sure, here's one way of explaining, which I hope is simple enough.
Firstly, we forget about the circle and only consider the arrangment
of n different objects in a straight line.
For 1 object A, there's only 1 way.
For 2 objects A, B there are 2 ways, ie. AB or BA, ie 2 ways = 2x1 ways = 2! ways.
For 3 objects A, B, C there are 6 ways = 3x2x1 ways = 3! ways. (Try to
list out all the 6 ways to confirm).
For 4 objects A, B, C and D there will be 24 ways = 4x3x2x1 = 4! ways.
(Again try to list out all the 24 ways to confirm).
In general then, no. of ways to arrange n different objects in a
straight line is n! ways.
You have to know the above result first before proceeding.

We now go to arranging n different objects in a circle. 
For 2 objects A, B in a circle, the arrangement AB and BA in a circle
are the same arrangement - ie. only 1 way to arrange 2 objects in a
circle. (Note that for the straight line, AB and BA are 2 different
arrangments).

For 3 objects A, B, C in a circle the arrangement ABC, BCA, CBA in a
circle is the same arrangement (1 way). Similarly, the arrangement
ACB, CBA, BAC in a circle is the same arrangement (1 way) and these
are the only arrangments. ie total 2 ways = (3-1)! for 3 objects in a
circle. (If you have listed the 6 ways = 3! for 3 objects in a
straight line, you will see that these 6 arrangements in a straight
line gives rise to only 2 different arrangments for 3 objects in a
circle).

Similary for 4 different objects in a circle, the arrangement ABCD,
BCDA, CDAB, DABC are the same arrangment in a circle, etc. Thus
instead of 4! arrangments in a straight line for 4 objects, we now
only have 4!/4 = 3! = (4-1)! arrangments for 4 objects in a circle,
since 4 of the different arrangments in the straight line are exactly
the same arrangement in the circle.

Hence for n different objects in a circle, the no. of arrangments is 
n!/n =(n-1)! arrangements as answered previously.

Further note:
Note that the results for the circle above apply only to circles which
cannot be flipped over, ie. they apply to arranging people in circle,
arranging plants around a circle, etc.
If a circle can be flipped over, like a bead, then the results is as follows:
No. of arrangments for n different objects in a bead is (n-1)!/2. The
explanation is similar to the above and will be skipped.

Hope the above explanation is simple enough. Guess I've saved you 2 bucks!

Great explaination pkuanko!!

Eric

one additional problem:two arrangements are considered the same if
each guest is seated between the same people.what is the number with
this added infomation. thank you.

As to your further refinement, I believe that would halve the number
of possibilities.  To use the letters as pkuanko did, with the new
refinement

ABCD
will be the same as
DCBA

I can't think of any other way to arrange them so each person is
between the same two (other than rotating them around, as pkuanko
already covered).

So if the original answer was 9!=362880, the new answer would be 9!/2=181440.

thank you!