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I know circular arrangement of $n$ different objects can be done is $(n-1)!$ ways.

For example :- I arranged $7$ objects in circle

This can be done in $720$ ways (using $6!$)

$1$) Can I also do this problem as below ??

I made circular arrangement of $6$ objects in $5!$ ways.

Then I selected $1$ gap out of $5$ gaps between objects to put $7$th object

$2$) If yes

then It will give answer = $5!$ $5$

=$600$ (not $720$)

my solution:-

($1$) it is possible

But gap should 6 (as there are 6 gaps if we arrange them in linear way)

($2$) we have to use different formula .

like $5!$ 6

But I am not sure

plz help

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rst
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    Your second version is correct: there are $6$ gaps, not $5$, so there are $6(6-1)!=6\cdot5!=6!$ ways to insert the $7$-th object into the circle of $6$ objects. This is of course the same as the original $(7-1)!$ calculation. – Brian M. Scott May 25 '13 at 04:30
  • @BrianM.Scott, thanks for the confirmation. – rst May 25 '13 at 07:46

1 Answers1

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Converting my comment to an answer to get this off the Unanswered list:

Your second version is correct: there are $6$ gaps, not $5$, so there are $6(6-1)!=6\cdot5!=6!$ ways to insert the $7$-th object into the circle of $6$ objects. This is of course the same as the result of the original $(7-1)!$ calculation.

Brian M. Scott
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