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There are 4 tasks to be allotted to students. Each task must be

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ChandlerBong
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There are 4 tasks to be allotted to students. Each task must be [#permalink] New post   27 Apr 2023, 00:35
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There are 4 tasks to be allotted to students. Each task must be allotted to exactly one of the 7 students: Albus, Betty, Cindy, Dudley, Emma, Fred, and Harry. If Harry doesn’t want to do Task 1 or Task 4 and Dudley does not want to do Task 2 or Task 3, in how many ways can we assign the task?

(A) 100

(B) 200

(C) 240

(D) 280

(E) 440
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E
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There are 4 tasks to be allotted to students. Each task must be [#permalink] New post   27 Apr 2023, 12:38
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ChandlerBong wrote:
There are 4 tasks to be allotted to students. Each task must be allotted to exactly one of the 7 students: Albus, Betty, Cindy, Dudley, Emma, Fred, and Harry. If Harry doesn’t want to do Task 1 or Task 4 and Dudley does not want to do Task 2 or Task 3, in how many ways can we assign the task?

(A) 100

(B) 200

(C) 240

(D) 280

(E) 440


To assign 4 tasks, we need four out of seven students. Therefore to complete the problem at hand, we need to perform the following two steps -

  1. Select a group of four students from seven students
  2. Assign tasks to each student within that group

We can create the group in four different ways -

  • Group 1: Harry is a part of the selected four but Dudley is not
  • Group 2: Dudley is a part of the selected four but Harry is not
  • Group 3: Harry and Dudley both are part of the selected group
  • Group 4: Neither Harry nor Dudley is a part of the selected group

Let's calculate the above two steps for each group

Group 1: Harry is a part of the selected four but Dudley is not

H _ _ _

Step 1: Select three students from available five students

We can select three students from the five available students in \(^5C_3\) ways

Step 2: Assign the tasks to each student

  • Harry can be assigned tasks in two ways (either task T2 or T3)
  • The remaining three students can be assigned tasks in 3! ways.

Total = \(^5C_3 * 2 * 3! = 120\)

Group 2: Dudley is a part of the selected four but Harry is not

D _ _ _

Step 1: Select three students from available five students

We can select three students from the five available students in \(^5C_3\) ways

Step 2: Assign the tasks to each student

  • Dudley can be assigned tasks in two ways (either task T1 or T4)
  • The remaining three students can be assigned tasks in 3! ways.

Total = \(^5C_3 * 2 * 3! = 120\)

Group 3: Harry and Dudley both are part of the selected group

D H _ _

Step 2: Select two students from available five students

We can select two students from the five available students in \(^5C_2\) ways

Step 2: Assign the tasks to each student

  • Harry can be assigned tasks in two ways (either task T2 or T3)
  • Dudley can be assigned tasks in two ways (either task T1 or T4)
  • The remaining two students can be assigned tasks in 2! ways.

Total = \(^5C_2 * 2* 2 * 2! = 80\)

Group 4: Neither Harry nor Dudley is a part of the selected group
_ _ _ _

Step 2: Select four students from available five students

We can select two students from the five available students in \(^5C_4\) ways

Step 2: Assign the tasks to each student

  • The tasks can be assigned without any restriction i.e. in 4! ways


Total = \( ^5C_4 * 4!\) = 120

Number of ways in which we can assign the task = 120 + 120 + 80 +120 = 440

Option E
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Re: There are 4 tasks to be allotted to students. Each task must be [#permalink] New post   28 Apr 2023, 12:12
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ChandlerBong wrote:
There are 4 tasks to be allotted to students. Each task must be allotted to exactly one of the 7 students: Albus, Betty, Cindy, Dudley, Emma, Fred, and Harry. If Harry doesn’t want to do Task 1 or Task 4 and Dudley does not want to do Task 2 or Task 3, in how many ways can we assign the task?

(A) 100

(B) 200

(C) 240

(D) 280

(E) 440


Quick prequel: a clarification is necessary on this question. As written, there is no stipulation that each task is done by a different student. Without that stipulation, the solution should be as follows.
We have 6 choices for who will do Task 1, 6 for Task 2, 6 for Task 3, and 6 for Task 4. That's 6^4 = 1296. Since that's not an option, let's assume that stipulation is in play and the question should be reworded.


Okay, how about a MUCH faster way?
(Sure the nPr/nCr stuff is great, but a little logic saves a lot of work...and GMAC doesn't care HOW you get the right answer, only THAT you get the right answer.)

Task 1 can be assigned to 6 people.
Once we've done that, we have 6 people left. If they include Dudley, then we have 5 options for Task 2. If they don't include Dudley, then we have 6 options for Task 2. So, either 5 or 6 options.
Once we've done that, we have 5 people left. If they include Dudley, then we have 4 options for Task 3. If they don't include Dudley, then we have 5 options for Task 3. So, either 4 or 5 options.
Once we've done that, we have 4 people left. If they include Harry, then we have 3 options for Task 4. If they don't include Harry, then we have 4 options for Task 4. So, either 3 or 4 options.

We know that the number of total options CAN'T be smaller than each of the smallest numbers from each row multiplied together. That's 6*5*4*3 = 360. Eliminate A, B, C, and D.

Answer choice E.
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Re: There are 4 tasks to be allotted to students. Each task must be [#permalink] New post   07 Apr 2024, 05:06
With no restrictions the tasks can be allocated:

7*6*5*4 = 840

Each has two tasks that can't be allocated to them, so the above needs to be reduced, with a preliminary approach as:

2*(3 remaining tasks among 6 people) =

2*6*5*4 = 240

Since this restriction applies to both Dudley and Harry:

2*240 = 480

Subtracting these restrictions from the initial 840:

840-480 = 360

Since we know there is overlap in the restrictions, some number would be added back.

The only answer that fits is:

440

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Re: There are 4 tasks to be allotted to students. Each task must be [#permalink] New post   07 Apr 2024, 06:35
Can it be solved faster?
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Re: There are 4 tasks to be allotted to students. Each task must be [#permalink]
07 Apr 2024, 06:35
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