Latest TCS Aptitude Question SOLUTION: 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.

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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

Asked In TCS Tejasree (10 years ago)

Unsolved Read Solution (11)

Is this Puzzle helpful?   (15)   (9) Submit Your Solution Permutation and Combination

Read Solution (11) :

sravya pusapati  (10 years ago)

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

Is this solution Helpfull? Yes (27) | No (4)  

Sruthi karuturi  (10 years ago)

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

Is this solution Helpfull? Yes (21) | No (15)  

Santosh Kumar  (10 years ago)

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

Is this solution Helpfull? Yes (7) | No (4)  

satyam  (9 years ago)

answer is 192

Is this solution Helpfull? Yes (3) | No (2)  

purnima  (10 years ago)

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

Is this solution Helpfull? Yes (2) | No (3)  

Nirmalya Basu  (9 years ago)

@purnima the question states "at least 1 task" so stupid comparison with seating arrangements, the answer is 192. Sruthi is correct.

Is this solution Helpfull? Yes (2) | No (3)  

Arpit Rawat  (10 years ago)

pls explain it clearly .......confused in sol

Is this solution Helpfull? Yes (1) | No (3)  

ashutosh  (9 years ago)

answer is 144

Is this solution Helpfull? Yes (1) | No (1)  

sujeeth s  (6 years ago)

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

Is this solution Helpfull? Yes (1) | No (1)  

sathiya.v  (9 years ago)

pls give clear solution

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satyam  (9 years ago)

sorry ans is 144

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