A teacher is assigning 6 students to one of threetasks. She will assig

Question

A teacher is assigning 6 students to one of three tasks. She will assign students in teams of at least one student, and all students will be as-signed to teams. If each task will have exactly one team assigned to it, then which of the following are possible combinations of teams to tasks?

I. 90
II. 60
III. 45

A. I only
B. I and II only
C. I and III only
D. II and III only
E. I, II, and III

I. 90
II. 60
III. 45

A. I only
B. I and II only
C. I and III only
D. II and III only
E. I, II, and III

CareerGeek · Accepted Answer

Each team gets at least 1 student
So possible combination of teams allotted = (4,1,1) or (3,2,1) or (2,2,2)

Number of ways to select (4,1,1) = 6c4*2c1*1c1 (select four out of 6, then select one out of 2 and select one out of 1)
= 15*2 = 30 ways

Number of ways to select (3,2,1) = 6c3*3c2*1c1 (select 3 out of 6, then select 2 out of 3 and select one out of 1)
= 20*3*1 = 60 ways

Number of ways to select (2,2,2) = 6c2*4c2*2c2 = (select 2 out of 6, then select 2 out of 4 and select 2 out of 2)
= 15*6*1 = 90 ways

So, Option B

Pls Hit kudos if you like the solution

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Archit3110 · Answer

for three tasks we have 6c2*4c2*2c2 /3! ways of distribution 
viz 15 ways
total tasks 3 which can be assigned in 3! ways ; 6 so total combination possible 15*6 ; 90
IMO A

nick1816 · Answer

There must be 3 teams and each team must have a students. Now there are 3 ways to divide 6 members in 3 teams.
1. When each team has 2 members.
Number of to assign students in 3 teams- 6!/(2!2!2!3!) =15
Number of ways to assign them 3 tasks = 15*3!=90
2. When two teams has 1 member and one team has 4 members
Number of to assign students in 3 teams- 6!/(4!1!1!2!) =15
Number of ways to assign them 3 tasks = 15*3!=90
3. When 1 team has 3 members, second team has 2 members and third team has 1 member.
Number of to assign students in 3 teams - 6!/(3!2!1!)= 60
Number of ways to assign them 3 tasks = 60*3!=360

akansal211 · Answer

Can someone provide solution for this.

Shishou · Answer

There are 3 tasks and 6 students. the students are to be divided into teams and each team should have at least 1 student.Each task should have a team attached to it. So we could divide the teams among tasks 1,2 and 3 into teams with members (1,3,2) or (2,2,2) or (4,1,1). To create teams of 1 person, 3 persons and 2 persons (1,3,2) gives :6C1*5C3*2C2. This gives us 60 ways. Then to create teams with members (2,2,2) we will get 6C2*4C2*2C2. This gives us 90. 
The final set will then be 6C4*2C1*1C1 which will be 30. Only 90 and 60 are represented in the options above.The answer is thus I and II which is B.

Fdambro294 · Answer

Bunuel

nick1816

ANY  expert who feels like chiming in lol

Am I approaching this question incorrectly?

We have distinct people ————> and we are distributing them to distinct “tasks”, not identical “stacks”

The “60” option everyone has above is assuming that we are just splitting the people into non-distinct teams of: 3, 2, and 1 —— and then not assigning the teams to the 3 different tasks.

First, we can start by choosing the people to place in “formations” or “piles” or whatever you want to call it (here “teams” is meant to be identical teams, not distinct....so teams is the right word I suppose)

“6 choose 3”

Then: of 3 remaining: “3 choose 2”

Finally, of 1 remaining “1 choose 1”

Works out to:

6! / (3! 2! 1!) = 60 ways to distribute the different people into “identical teams” totaling 3 people, 2 people, and 1 person

Now for each 1 of these 60 possible apportioning of teams, we can shuffle among the 3 DISTINCT TASKS in 3! Ways

60 * 3! = 360 ways

The possibilities are:

90, 90, and 360

Where have I made a wrong turn?

Any help would be greatly appreciated.

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Fdambro294 · Answer

Ah, let me try

chetan2u

I believe you wrote a whole article on distributing distinct items to distinct groups and the other various forms.....(well-written)

The way the question is written, my interpretation was the following:

First, the teacher is assigning the 6 students to 3 teams. Each team must have at least one student, so there are 3 unique ways to break up the students by number:

For these 3 unique cases of # of ppl per team, we first must find the number of ways to divide the 6 students into 3 teams

Second, we are given and told that “each task will have one team assigned to it.” I read this to mean that once we have determined all the possible ways to divide the students among 3 teams, we must then figure out all the different ways we can assign each team to cover the 3 distinct tasks. (“Shuffling” the 3 teams among the 3 distinct tasks)

3 distinct teams assigned to 3 distinct tasks ——-> 3! = 6 Ways to arrange the teams to the different tasks

Breaking up the students into teams containing at least one student, what follows is the calculation for each case.

Case 1:

(1st) break up/divide the 6 students into 3 teams that have the following sizes

—-1
—-1
—-4

First, from the 6 students we must choose 4 for one of the teams——> this can be done in “6 choose 4” ways

From the 2 remaining students, we choose 1: “2 choose 1”

Finally, the last one is automatically chosen from the 1 remaining student: “1 choose 1”

*   *   =

After you cancel Factorials in the NUM/DEN across the expressions, you are left with:

6! / (4! 1! 1!)

However, we have 2 teams that have an identical number of people. We will be over counting as if they are being distributed into distinct teams, when, at this point, we should only be dividing the students into “identical groups.”

Label the 6 students A through F. We will have included in our “over-count” examples such as:

———  ———-

and

——— ———-

Both will be included in the “over-count”, but for purposes of first dividing the students into “identical” teams, these 2 cases should only be counted ONCE

For every 2 cases we counted in our over-count, we only want to actually count 1

We must divide our expression by (2!) to remove this over-counting (now for every 2 cases counted, only 1 will remain)

6! / (4! 1! 1! ) * (1 / 2!) ———-> which is the same as “6 choose 4” ———-> 
6 * 5 / 2 =

15 ways to divide the students into teams of sizes

AND

(2nd)for each one of these 15 ways to create teams, there will be 3 teams of students. We want to determine how many ways we can arrange these 3 teams among the 3 distinct tasks. This can be done in:

3! = 6 ways

15 * 6 = 90 total ways for case 1

Case 2: for the next 2 cases I will just perform the calculations. The logic is similar.

(1st) # of ways to divide students into teams of

6! / (3! 2! 1!) = 60 ways

AND

(2nd) for each one of these 60 ways of dividing students into teams of sizes 3, 2, and 1, we need to arrange or “shuffle around” the 3 teams among the 3 distinct tasks. We can do this in:

3! = 6 ways

Case 2 total no. of ways:
60 * 6 = 360 ways

*note*** this is the case in which the other answers are coming up with the 60 possibility. The students are divided among 3 “identical groups” of teams, but then the teams are not arranged among the 3 distinct tasks.

Case 3:

(1st) find the # of ways to divide the 6 students into 3 teams of size 2, 2, and 2. Since we have Three teams that have the same number of students, we will have the same over-counting issue explained in case 1. Among the counted examples:

-   -  
  -   -  
  -   -  
etc.

For purposes of dividing the students into identical teams, each of the examples above are the same and should only be counted once. But for each once of these cases, we have over-counted and actually included 3! = 6 ways

This time, we need to divide by (3!) —- for each 6 cases included in the over-count, we only went to keep 1 in the final count.

6! / (2! 2! 2!) * (1 / 3!) = 15 ways

AND

(2nd) for each of the 15 possibilities of dividing the students into teams, there will be 3 teams of sizes 2, 2, and 2. We need to arrange these 3 teams among 3 distinct tasks. We can do this in:

3! = 6 ways

Case 3 total:
(15) * (6) = 90 ways

Possibilities:
90 ways (case 1)

Or

360 ways (case 2)

Or

90 ways (case 3)

The answer should be A. I only

Am I wrong? Could you kindly point me to where I made a wrong turn?

Thank you for any help.

All the best!

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MathRevolution · Answer

There are two possible answers of this question as the sentence structure is not proper.

We need to take at least 1 student in each team to form three teams.
We have a total of 6 students at our disposal and three tasks.

Various possibilities to assign students:

Case I:  2, 2, 2: ^6{C_2} * ^4{C_2} * ^2{C_2} = 15 * 6 * 1 = 90  - Option I

Case II:  3, 2, 1: ^6{C_3} * ^3{C_2} * ^1{C_1} = 20 * 3 * 1 = 60  - Option II

Case III:  4, 1, 1: ^6{C_4} * ^2{C_1} * ^1{C_1} = 15 * 2 * 1 = 30  - Option not there.

So, the answer is B.

NOTE:

But in Case II: The given statement in question is Each task will have exactly 1 team assigned. So, which task will have a team of 3, a team of 2, and a team of 1 has to be assigned first.

So select 3 students out of 6 and at the same time select 1 task out of 3:  ^6{C_3} * ^3{C_1} = 20 * 3 = 60 .

Simialrly, select 2 students out of 3 and at the same time select 1 task out of 2:  ^3{C_2} * ^2{C_1} = 3 * 2 = 6

Lastly, 1 student is left for 1 task: 1

Total ways: 60 * 6 = 360 - OPTION NOT THERE.

Consider the same for Case III: 4, 1, 1

Select 4 students out of 6 and at the same time select 1 task out of 3:  ^6{C_4} * ^3{C_1} = 15 * 3 = 45 .

Simialrly, select 1 students out of 2 and at the same time select 1 task out of 2:  ^2{C_1} * ^2{C_1} = 2 * 2 = 4

Lastly, 1 student is left for 1 task: 1

Total ways: 45 * 4 = 180 - OPTION NOT THERE.

Hence, the answer should be A

Fdambro294 · Answer

Thank you for the response  MathRevolution  and  nick1816

Something seemed off.....

The question goes out of its way to mention that the teams will be assigned to 3 unique tasks (there is no indication that the tasks are considered “identical” for purposes of the question)

So in the end, we are distributing distinct items (people) —————> to distinct groups of different sizes.

The official answer that is posted is correct if and only if we consider “identical” the tasks to which the teams of students are assigned.

In other words, the official answer works only if we believe any one of the tasks a particular student performs as part of a team is identical to the other two tasks.

If the students were chocolates instead of human beings, then we would be dividing the chocolates into “identical stacks.”

This does not appear to be what the syntax of the question conveys.

I just have one further question:

For case number 3 written in red in which 180 is found as a possible answer, wouldn’t you have to account for the fact that TWO of the teams are of identical size -  ?

Thus, in choosing the students, we will end up with an over count. For instance, if the students are: A, B, C, D, E, F———> a count of 180 would include both:

——-   ————

And

———   ————

For purposes of dividing the students into teams, both distributions would be the same. Then, for each one of the unique distributions, we can arrange the 3 teams to the 3 distinct tasks (which we can do in 3! Ways).

Thus, for every 2 distributions of students that we included in the 180 count, we should only actually count 1.

Thus multiply 180 by:

(180) * (1 / 2!) = 90

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bumpbot · Answer

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