Question 31493
9 ARE TO BE ON 1 SIDE.4 ARE ALREADY ON ONE SIDE AND 3 ON OTHER SIDE.
THERE ARE 11 LEFT. 
WE CAN SELECT 5 OF THEM IN 11C5 WAYS TO SIT ALONG WITH 4.

REST WILL AUTOMATICALLY GO TO OTHER SIDE.THIS CAN BE DONE IN ONLY 1 WAY .
NOW 9 ON EITHER SIDE CAN BE ARRANGED IN 9! WAYS...SO THE TOTAL NUMBER OF ARRANGEMENTS ARE 
11C5*9!*9!.....A IS THE ANSWER