2
$\begingroup$

If 20 persons were invited for a party, in how many ways will two particular persons be seated on either side of the host in a circular arrangement?

According to me the answer should be $17!.2!$. But the given answer is $18!.2!$.

If we consider the guest and the host as one unit and let them take the first three chairs the other 17 can be occupied by 17! ways and the two particular persons can then rearrange them by 2! ways.

What am i doing wrong?

$\endgroup$

1 Answer 1

Reset to default
2
$\begingroup$

when you treat the two guests and the host as one unit then you are left with $19$ entities to arrange ($18$ remaining guests $+$ this combined unit). The number of ways to arrange them is $18!$ and then you can permute the two special guests in $2!$ ways. In all $18! 2!$.

$\endgroup$
3
  • $\begingroup$ If I am left with 18 entitities should not i be making (n-1)! =(18-1)! arrangements? This is circular permutation problem $\endgroup$
    – archangel89
    May 4, 2015 at 7:23
  • $\begingroup$ @archangel89 You are right about circular arrangement and I had factored that but unfortunately I typed it incorrectly. Note there are $20$ guests and $1$ host so in all $21$ people. Now see the solution (I have edited it) $\endgroup$
    – Anurag A
    May 4, 2015 at 7:28
  • 1
    $\begingroup$ Oh thank you. I forgot host is different from the guests. $\endgroup$
    – archangel89
    May 4, 2015 at 7:29

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .