Six persons  $A,B,C,D,E$  and F are to be seated at a circular table facing towards the centre then the  number of ways that can be done if A must have either E or F on his immediate right and E must have either F or D on his immediate right, is.
 

Case- I : E is to the immediate  right of A
               The seat next to E (on right side) can be filled in 2-ways
               i.e., either D or F remaining 3-seats can be filled in 3! Ways
               Total no. of ways =  $2\times 3!=12$
Case- II : F is to the immediate right of A number of ways E can be filled = 3-ways
               [ E can’t occupy seat on left of A as there will be no Space for D]
               Number of ways for D = 1 ways
               Remaining seats can be filled in 2 –ways.
               Total number of ways  $=3\times 1\times 2=6$
                required total number of ways  $=12+6=18$