18 guests have to be seated,half on each side of a long table .4 particular guests desire to sit on one particular side and 3 others on the other side .Determine the number of ways in which the seating arrangement can be made

• 9p4*9p3*11!*2
  4 guests one side out of 9 seats =9p4
  3 guests another side out of 9 seats =9p3
  remain 11 guest can sit in 11 seats any where =11!
  the 3 and 4 perticular guest can change their side =2
• 12 years agoHelpfull: Yes(39) No(13)
• 9p4*9p3*11!
• 12 years agoHelpfull: Yes(28) No(8)
• 9p4*9p3*11!
• 12 years agoHelpfull: Yes(11) No(3)
• 11c5*6c6*9!*9! (ans)
  
  The reason is simple, for the first table 4 particular people sits together and 3 for the second. Therefore 11 people out of 18 are left. so for the first table they can be selected in 11c5 ways. Similarly for the second table the no of combination is 6c6. Now the people can be arranged among themselves in 9!*9! ways.
  Therefore total no. of arrangements=11c5*6c6*9!*9!
• 12 years agoHelpfull: Yes(11) No(2)
• 6!*4!*7!*3!*2
  out of 9, 4 sit together.so total number of units are 6.(9-4=5 and 4 together as 1 unit).so they can be arranged in 6!*4! ways...similarly on another side, if 3 sit together 7!*3! ways. and 2! is for if they select an other side.
• 12 years agoHelpfull: Yes(7) No(2)
• 9p3*9p4*11!
• 12 years agoHelpfull: Yes(3) No(2)
• 11 C 5 * 9! * 9 !......................................D
  
  side 1
  ______________
  
  side 2
  
  
  18 guests
  4 chooses to sit on side 1
  3 chooses to sit on side 2
  
  now 11 left
  now 14 guest left
  we have to choose 5 guests for side 1 in 11 C 5 ways
  rest 6 will seat on other side
  and now each side can be arranged in 9! ways
  hence total arrangement = 11 C 5 * 9! * 9!
• 10 years agoHelpfull: Yes(2) No(0)
• bro 11! is in which manner come pls explain
  
• 12 years agoHelpfull: Yes(1) No(1)
• 9p4*9p3*11!*2
  
• 12 years agoHelpfull: Yes(1) No(4)
• 9p4*9p3*11!
• 11 years agoHelpfull: Yes(1) No(1)
• !6*!4*!8*!3
• 12 years agoHelpfull: Yes(0) No(9)
• 9c7*9c6*4!*3!
  9 on each side.
  4 persons fixed so tak dat to b 1.
  der fore no of ways is 9c6 ways * 4!
  same on the odr side 9c7 * 3!
• 12 years agoHelpfull: Yes(0) No(11)
• at first take 4 guests in one side and likewise take 3 guests for other side.now 11 guests remains.so for first side where only 9 guests are allowed the remaining 5 guests will be taken out of 11 in C(11,5) ways , similarly for 2nd side where also 9 guests are allowed the remaining 6 guests will be taken out of 11 in C(11,6) ways.
  in every side the guests can be arranged in 9! ways .
  hence the solution is C(11,5)*C(11,6)*9!*9! ways=2.810*10^16
• 12 years agoHelpfull: Yes(0) No(2)
• 9p4*9p3*11!*2
• 10 years agoHelpfull: Yes(0) No(3)
• 11c5*9!*9!+11c6*9!*9!
  =11c5*9!*9!+11c5*9!*9!
  =2*11c5*9!*9!..
• 9 years agoHelpfull: Yes(0) No(0)