Question

If $12$ persons are to be seated around a circular table such that the two secretaries of the chairman should be seated adjacent to the chairperson on both the sides and three particular boys should never seat together, then the total number of arrangements will be

• A. $66(7!\times 2!)$
• B. $135\times 7!$
• C. $72(7!\times 2!)$
• D. $108\times 9!$

Answer 1

The correct option is A $66(7!\times 2!)$

total number of arrangements when two secretaries of the chairman seats adjacent to the chairperson = $9!\times 2!$ ($\because$ secretaries can exchange their places)
In the above case if we subtract the arrangements when three boys always seats together then we get the required number of arrangements
$9!\times 2!-7!\times 2!\times 3!=66(7!\times 2!)$