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# Factorial

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Manager
Joined: 24 May 2010
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24 May 2010, 22:35
Ok so if 5 people are to sit at a round table how many ways can they be seated. Why is the answer not 5!
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25 May 2010, 00:07
Total No. of ways in which n no. of persons could be arranged on a round table is given by : ( n - 1 ) ! and not n !

Therefore the ans shd be 4! and not 5!
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25 May 2010, 07:03
Yes but why n-1 ! And not n! Can you give some more color so I can understand

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25 May 2010, 07:33
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Jinglander wrote:
Yes but why n-1 ! And not n! Can you give some more color so I can understand

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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 $$(n-1)!$$.

From Gmat Club Math Book (combinatorics chapter):
"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. So, for the number of circular arrangements of n objects we have:

$$R = \frac{n!}{n} = (n-1)!$$"

$$(n-1)!=(5-1)!=24$$

Check Combinatorics chapter of Math Book for more (link in my signature).

Hope it helps.
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15 Feb 2016, 03:22
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26 Feb 2016, 17:21
In a circle there are n-1 ways to arrange a group
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28 Feb 2016, 22:32
One way to think about it, and these problems in general, is to start by counting the possibilities 'naively'. It seems logical that there should be 5! ways to arrange 5 people around a round table, so start with 5!. Then, account for any special circumstances by figuring out whether you actually counted any of the possibilities more than once. In this case, you actually counted each separate possibility five times by using 5!. For instance, you counted these five arrangements as being different, but they're actually the same (since they're just rotations around the table):

(A B C D E)
(B C D E A)
(C D E A B)
(D E A B C)
(E A B C D)

In order to correct for the overcounting, you'll divide by 5. 5!/5 = 4!, or 24.

This works for a wide range of counting problems. Suppose you wanted to know how many ways a class of eight people could be split into two groups of four. Naively, there are 8*7*6*5 ways to select the first group of four (after which the second group is determined). But you've overcounted by doing that, since you actually counted these groups as being different:

A B C D
B C D A
B A C D
... etc.

That is, you counted each different group 4! times. So, the actual answer is (8*7*6*5)/4!, which is equivalent to what you'd get from the combinatorics formula.
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Re: Factorial   [#permalink] 28 Feb 2016, 22:32
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