18 guests have to be seated half on each side of a long table.four particular guests desire to sit on one particular and three others on other side.the no. of ways in which sitting arrangement can be made?

• four on one side and three on other side.. hence guests number is limited to 11 as 7 persons are already fixed, since 18-(4+3)=11.
  --->now consider the side with four people, hence we have to fill five more places,
  selection= 11C5 arrangement= 9! (since the guests are fixed to only sides not the places on that particular side)
  ----> consider the other side we already have 3 people fixed and also remaining are 6 guests as already five are selected before for other side.
  hence only arrangement= 9!
  
  total= 11C5*9!*9!
• 12 years agoHelpfull: Yes(19) No(4)
• (2c1)*(9p4)*(9p3)*(11p1)
  
  =============>
  2c1--->select a side
  9p4------>4people can be arranged in 9 seats(one side) in 9p4 ways
  9p3------>3people can be arranged in 9 seats(other side) in 9p3 ways
  11p11----->remaining 11 people can sit any where so 11 people in 11 seats therefore 11p11 or 11!
• 12 years agoHelpfull: Yes(9) No(2)
• the 4 person on one side can be seated in 9C4*4! ways.
  the 3 person on the other side can be seated in 9C3 *3! ways.
  now the two sides has to be multiplied by 2!
  the remaining 11 person can be seated in 11! ways...
  
  so total = 9C4*4!*9C3*3!*2*11!....
• 12 years agoHelpfull: Yes(4) No(4)
• 11c5*9!*9!*2 is the answer
• 12 years agoHelpfull: Yes(3) No(0)
• 3 persons will sit on 1 side and 1 person on other side.
  now we have to select 6 persons for 1st side=14C6(which is equal to 14C8 for other side)
  now on both sides 9 persons can be arranged in=9! ways
  so total no. of ways= 14C6*9!*9!
• 12 years agoHelpfull: Yes(1) No(3)
• IF IT IS A RECTENGLAR TABLE THEN IT CAN BE SOLVED AS FOLLOWS
  FIRST 9 PERSONS WILL SIT ON EACH SIDE OF TABLE
  THEN SELECT OF 9 PERSONS FROM 18 PERSONS = 18C9 WAYS
  NOW EACH 9 CAN BE SEATED AS= 9! X 9!
  AFTER REST 4 GUESTS CAN BE SATED AS BETWEEN THE SPACE CREATED AMONG THESE PERSON
  OUT OF IT ONE PERSON WILLING TO SIT ON A PARTICULAR SPACE SO REMAINING SPACES=21
  SO THIS CAN BE DONE = 18C9 x 9! X 9! x 21C3 (ANS)
• 12 years agoHelpfull: Yes(0) No(19)
• if it is a rectangular table then each side 9 guests can sit.
  on one side of table always 4 particular guest will sit together so they can be sit in=6!*4P1 WAY.
  ON the otherside always 3 will always sit together so the can be sit in=7!*3P1 WAY
  SO ALLTOGETHER THEY CAN SIT IN=(6!*4P1)*(7!*3P1)WAY=43545600WAY
• 12 years agoHelpfull: Yes(0) No(5)
• 4!*3!*11P5*11P6
• 12 years agoHelpfull: Yes(0) No(0)
• 9p4*9p3*11
• 12 years agoHelpfull: Yes(0) No(0)