In how many ways can the entrepreneur assign 5 different tasks to 3 employees if each should get atleast 1 task ?

Question

For this I considered cases

1. 1 job each to 2 people and then  jobs to a single person

2. 2 jobs each to 2 people and then 1 job to a single person

For the first case I did

5C1*4C3*1C1 ,Now this can be permuted in 3! ways

for the second case I did 5C2*4C2*1C1 , Again this can be permuted in 3! ways ,Now on adding these two cases I am getting answer as 480 , what's wrong in this approach ?

Comments

this problem can be converted into no of onto function from set containing 5 element to set containing 3 element

then you get 150 answer

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in a simple way 

 A  B   C

1   2   2

2   1   2

2   2   1

3   1   1

1   1   3

1   3   1

 

Answer 1

Let the tasks employees be numbered E1, E2 and E3.

As every employee is given at least 1 task, we assign them 3 out of 5 tasks. We are left with 2 tasks, name them A and B.

Now, we have to assign these 2 tasks: A and B, to 3 employees : E1, E2 and E3.

Each task can be assigned to any of the 3 employees, we don't care if an employee gets both of these 2 tasks, or gets none of these 2, because we have already assigned 1 task to each employee and we can assign the remaining 2 tasks in any way we wish.

So, each task can be assigned to any of 3 employees. 2 tasks can be assiged in 3*3=9 ways.

Comments

hi,

couldnt this be done with no of onto functions possible ?? keeping m as 3 and n as 5.

5³-5c1(4)³+.......... ??

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this shouldn't be multiplied as 3*3... this must be add up as 3+3=6 ways.. coz for first task there are 3 options and for second task 3 options..

Answer 2

The assignment is similar to onto function fron set of task to set of employees.

so the number of onto function are= 3^5-3C1 (2^5)+3C2 (1^5)= 150.