There are 6 people at a party sitting at a round table with 6 seats: A, B, C, D, E and F. A CANNOT sit next to either D or F. How many ways can the 6 people be seated?  Choose the most appropriate option:  A) 720  B) 120  C) 108  D) 84  E) 48
A,B,C,D,E and F can sit in (6-1)! ways -- 120   Now consider this, when we take AD as one person all of them can sit in AD, B, C, D ,E --&gt; (5-1)! --&gt; 24   now, this calculation considers clockwise and anti clockwise as different, i.e AD is different from DA. Therefore we have to divide this figure by 2, because We do not want to distinguish between AD and DA.   Hence there are 12 cases when AD sit to each other ( irrespective of AD or DA )   Similarly, there are 12 cases when AF sit to each other.   Similarly, we can say ADF can also be clockwise and anticlockwise, but we have to realize that, when counting AD and AF we have counted DAF or FAD TWICE ( once for each ) therefore we need to deduct the arrangement for DAF only once.   If DAF were 1 person, possible number of arrangements -- ADF,B,108,E --&gt; 3!   We will not count it twice because we need to deduct this only once from AD and AF combination !   Therefore total count   5! - (4!/2 + 4!/2 -3!) = 120 - (12 + 12 -12 )= 
<gadget id="calculator">120 - (12 + 12 -12 )</gadget>
<output>108</output>
108   ANS:108
<result>C</result>
There are 6 people at a party sitting at a round table with 6 seats: A, B, C, D, E and F. A CANNOT sit next to either D or F. How many ways can the 6 people be seated?