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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 2must 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?

(A) 144
(B) 189
(C) 192
(D) 368
(E) 378

As TASK 2 can be assigned to either person 3 or 4 ,let's start with it.

Let task 2 goes to person 3 .

Then choices left for task 1 are Person 4,5,6 = 3

Rest of the tasks can be assigned in a non-restrictive way (4 task remaining and 4 person= 4! or 4 *3*2).

so total for this case = 3 * 4*3*2 = 72

Similarly 72 when task 2 goes to person 4 .

So total = 144 .Option A
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JohnGalt1
Pretty Sure its wrong but, lets give a try

_ _ _ _ _ _ => 192 => Thus, C.
4 2 4 3 2 1

I guess for this question you need to split the sum into cases. The first task and second task can have conflicting numbers.

JohnGalt1 Please have a look.
Task 1: has 3 possibilities because Task 1 cannot be assigned either to person 1 or to person 2 and at the same time task 2 is assigned to either person 3 or person 4.
Task 2: has 2 possibilities because task 2 must be assigned to either person 3 or person 4
Task 3: has 4 possibilities because remaining person are 4
Task 4: has 3 possibilities because remaining person are 3
Task 5: has 2 possibilities because remaining person are 2
Task 6: has 1 possibilities because remaining person are 1
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JohnGalt1
Pretty Sure its wrong but, lets give a try

_ _ _ _ _ _ => 192 => Thus, C.
4 2 4 3 2 1


Think this is wrong because Task 1 and task 2 have related people so you cannot multiply like that

Suppose for the first task we choose Number 3 person then would we have 2 options for the second task? No.

Answer is 144.
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mukherjeeabhish
There are 6 tasks and 6 persons. Task 1 cannot be assigned either to person 1 or to person 2; task 2must 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?

(A) 144
(B) 189
(C) 192
(D) 368
(E) 378

As TASK 2 can be assigned to either person 3 or 4 ,let's start with it.

Let task 2 goes to person 3 .

Then choices left for task 1 are Person 4,5,6 = 3

Rest of the tasks can be assigned in a non-restrictive way (4 task remaining and 4 person= 4! or 4 *3*2).

so total for this case = 3 * 4*3*2 = 72

Similarly 72 when task 2 goes to person 4 .

So total = 144 .Option A

I agree with your approach

Task 2 can be assigned in 2 ways

The previous assignment restricts the possibilities for task 1 to 3

Finally, there are 4 people that can be assigned to 4 tasks, thus ----> 2 * 3 * 4! = 6 * 4 * 3 * 2 = 144 -----> A
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