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Let T1,T2,T3,T4,T5 &T6 be the Tasks and P1,P2,P3,P4,P5 & P6 be the Persons

Task T2 can be done only in 2 ways (either to P3 or P4)

Task T1 can be done in 3 ways (P5,P6, and 1 among P3/P4)

Task T3 can be done in 4 ways (Remaining 4 people)

Task T4 can be done in 3 ways (Remaining 3 people)

Task T5 can be done in 2 ways (Remaining 2 people)

Task T6 can be done in 1 way (Remaining 1 person)

So, totally there can be a combination of 2 * 3 * 4 * 3 * 2 * 1 ways = 144 Ways.

Hence option A is the answer
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Task 1 to person 3 > Task 2 to person 4:
1*1*4*3*2*1 = 24

Task 1 to person 4 > Task 2 to person 3:
1*1*4*3*2*1 = 24

Task 1 to person 5 or 6:
2*2*4*3*2*1 = 96


Total: 24+24+96 = 144
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There are 6 tasks and 6 persons. Task 1 cannot be assigned either to person 1 or to person 2; Task 2 must be assigned to either person 3 or person 4. Every person is to be assigned one task. In how many ways can the assignment be done?

Combination # 1:
T1 -> P3 , T2 -> p4 then T3 will have 4 options, T4 3 options, T5 - 2 options and T6 - 1 option
1*1*4*3*2*1 = 24

Combination # 2:
T1 -> P4 , T2->P3, then others will have same options as above
1*1*4*3*2*1 = 24

Combination # 3:
T1 is assigned to person 5 or person 6.
T2 can be assigned to person 3 or person 4.
2*2*4*3*2*1 = 96

Adding all together = 24+24+96 = 144

Ans: A
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Kritesh
There are 6 tasks and 6 persons. Task 1 cannot be assigned either to person 1 or to person 2; Task 2 must be assigned to either person 3 or person 4. Every person is to be assigned one task. In how many ways can the assignment be done?

(1) 144

(2) 180

(3) 192

(4) 360

(5) 716
­6 tasks so T1, T2, T3, T4, T5 and T6.
6 person so P1, P2, P3, P4, P5 and P6.

Given, T2 can be done by either P3 or P4 i.e. 2 ways.
Also given, T1 can be done by T5, T6 and the remaining of P3 or P4 i.e. 3 ways.

Now calculate the other tasks: T3 - select 1 of the remaining 4 persons i.e. 4 ways. Similarly T4, T5 and T6 in 3, 2 and 1 ways.

Total: 2*3*4*3*2*1=144 ways. Option (A) is correct.
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