3. 6 task and 6 persons. P1 and P2 does not do task T1. T2 is assigned to P3 or P4;. Each person should be assigned with at least 1 task. In how many ways the task can be assigned.
a. 192
b. 360
c. 144
d. 180

• t1,t2,t3,t4,t5,t6
  t1 can be arranged in 3 ways(except p1,p2,(p3 or p4)) rem 3 ways
  t2 can be two ways(p3 or p4)
  and rem in 4! ways
  ans is 3*2*4!=144
• 10 years agoHelpfull: Yes(27) No(4)
• 192
  As T1 can be done in 4 ways(p3,p4,p5,p6)
  T2 in tw0 ways (p3,p4)
  and the remaining in 4! ways
  4*2*4!=192
  
• 10 years agoHelpfull: Yes(21) No(15)
• T1 T2 T3 T4 T5 T6
  3(ways) P3 (4! w a y s)=3*4!=72
  3(ways) P4 (4! w a y s)=3*4!=72
  ans=144
• 10 years agoHelpfull: Yes(7) No(4)
• answer is 192
  
• 10 years agoHelpfull: Yes(3) No(2)
• consider it as linear seating arrangement....i.e consider tasks as seats and 6 persons had to sit accordingly....
  _ _ _ _ _ _
  in 1st place only p5,p6,p4/p3 can sit...3c1
  in 2nd place only p3/p4 sits.....2c1
  in 3rd,4th,5th,6th places remaing 4 sits in 4! ways...
  =>3c1*2c1*4!=144
• 10 years agoHelpfull: Yes(2) No(3)
• @purnima the question states "at least 1 task" so stupid comparison with seating arrangements, the answer is 192. Sruthi is correct.
• 10 years agoHelpfull: Yes(2) No(3)
• pls explain it clearly .......confused in sol
  
• 10 years agoHelpfull: Yes(1) No(3)
• answer is 144
  
• 10 years agoHelpfull: Yes(1) No(1)
• T1 T2 T3 T4 T5 T6
  P1 P2 P3 P4 P5 P6
  In T1 does not exist P1 and P2=P3 P4 P5 P6.............4 WAYS,,,,here P6 is placed
  In T2 does exist P3 P4...........2 WAYS,here if P3 is placed
  T3 remaining P4 P5 P1 P2.......4 WAYS,HERE P4 IS PLACED
  T4 REMAINING P5 P1 P2.......3 WAYS,HERE P5 IS PLACED
  T2 REMAINING P1 P2........2 WAYS,HERE P1 IS PLACED
  T1 REMAINING P2.......1 WAY
  4*2*4*3*2*1=192 IS THE CRT ANSWER
• 7 years agoHelpfull: Yes(1) No(1)
• pls give clear solution
• 10 years agoHelpfull: Yes(0) No(1)
• sorry ans is 144
• 10 years agoHelpfull: Yes(0) No(1)