Question

A team of $8$ persons including ''Captain'' and ''Vice-captain'' are to be seated around a circular table, then the number of possible arrangements

• A. without any restriction is $7!$
• B. when ''Captain'' and ''Vice-captain'' should be seated opposite to each other is $5!$
• C. when there should be atleast one person between ''Captain'' and ''Vice-captain'' is $5\times 6!$
• D. when there should be exactly one person between ''Captain'' and ''Vice-captain'' is $12\times 5!$

Answer 1

Answer: (A), (C), (D)

The correct options are
A without any restriction is $7!$
C when there should be atleast one person between ''Captain'' and ''Vice-captain'' is $5\times 6!$
D when there should be exactly one person between ''Captain'' and ''Vice-captain'' is $12\times 5!$

Number of ways to arrange $n$ persons in circular table is $(n-1)!$ ways
$\therefore$Total ways without any restriction is $(8-1)!=7!$

When ''Captain'' and ''Vice-captain'' should be seated opposite to each other:
Let ''Captain'' select the seat, it can be done in $1$ way (because it is circular table)
''Vice-captain'' can select the seat in $1$ way opposite to ''Captain''.
Now the remaining persons can be arranged in $6!$ ways.
$\therefore$Total ways are $1\times 1\times 6!$

When there should be atleast one person between ''Captain'' and ''Vice-captain'':
Total ways when ''Captain'' and ''Vice-captain'' should be seated together $=6!\times 2!$
Total ways $=7!-6!\times 2!=5\times 6!$

When there should be eaxctly one person between ''Captain'' and ''Vice-captain'':
The person can be selected in${}^6{C}_1=6$ ways.
The remaining arrangements can be done in $(8-3+1-1)!=5!$ ways.
While ''Captain'' and ''Vice-captain'' can interchange among themselves.
$\therefore$Total ways are $6\times 2!\times 5!$