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8 votes
8 votes
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 ?
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6 Answers

Best answer
10 votes
10 votes

Answer is 150

U made the cases correct , but in calculation u made some error, Let's examine - 

Let's take 3 employees as A,B,C 


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

     1  job to A , 3 jobs to B, 1 Job to C =  5C1 * 4C3 * 1C1 = 20 ways

     1  job to A , 1 job to B, 3 Jobs to C =  5C1 * 4C1 * 3C3 = 20 ways

     3  jobs to A , 1 job to B, 1 Job to C =  5C3 * 2C1 * 1C1 = 20 ways

                 So, total 60 ways {20+ 20 + 20}

Note: Here u took 5C1*4C3*1C1   * 3! which would be 120 {wrong }


Case 22 jobs each to 2 people and then 1 job to a single person

         2  jobs to A , 2 jobs to B, 1 Job to C =  5C2 * 3C2 * 1C1 = 30 ways

         2  jobs to A , 1 job to B, 2 Jobs to C =  5C2 * 3C1 * 2C2 = 30 ways

         1  job to A , 2 jobs to B, 2 Jobs to C =  5C1 * 4C2 * 2C2 = 30 ways

                         So, total 90 ways for case 2..


    Total ways = case 1 + case 2 = 60  + 90 = 150 (Ans)

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7 votes
7 votes

@radha gogia, your method is also right but u did a slight mistake there.

U observe that there are two patterns $3-1-1\;and\; 2-2-1$.
Now take first pattern, The first pattern can be chosen in $\binom53\binom21 = 20$ ways, that is exactly same as $\frac{5!}{3!1!1!} = 20$. Now u see how many ways you can allot, that is  $\frac{3!}{2!} = 3$ (NOT $3!$, u made it $3!$ and that's where u went wrong. this is similar to arranging three numbers $3-1-1$ in which 2 number are same, see mississippi )
For first pattern, I can write-  $\frac{5!}{3!1!1!} \times \frac{3!}{2!} = 60$ (these are total ways of choosing and then allotting )
Similarly for 2nd pattern, i can write $\frac{5!}{2!2!1!} \times \frac{3!}{2!} = 90$
Total = $60+90 = 150$

4 votes
4 votes

For example let set of tasks be {a, b, c, d, e} & set of employees be {x, y, z}, & consider following 2 cases:

Case - 1

From 5C3*3! we get

a --> x

b --> y

c --> z,

that is a, b, & are assigned to x, y, & z respectively.

& then both d & e assigned to x.

Case - 2

From 5C3*3! we get

b --> y

c --> z

d --> x

& then both a & e assigned to x.

there will be many such repetitions.


It can be observed that assigning tasks in rounds (as we did here assigned 3 tasks in round one & remaining 2 in round two)  will not work we should assign all at once.

This can be done correctly by breaking set of tasks into 3 partitions & assigning each partition to one employee.

So number of ways = (number of possible 3 partitions of set "Tasks")*3!

Counting number of partitions:

we can 3 - partition the set {a, b, c, d, e} into two different ways:

1) Partition of type 3, 1, 1 : Choose any 3 elements & put them in one partition. 5C3*1*1 = 10 partitions of this type.

2) Partitions of type 2, 2, 1 : Choose any 2 elements & put them in one partition, then choose 2 out of remaining 3 elements & put them into second partition, remaining one element will go to third partition without any choice. ((5C2)*(3C2)*1)/2 = 15 possible partitions of this type.We divided by two because all partitions of type 2, 2, 1 will be counted twice.

For example : suppose a partition {a, b}, {c, d} {e}, this will be produced in two ways, first choose {a, b} using 5C2 & {c, d} using 3C2 and another way choose {c, d} using 5C2 & {a, b} using 3C2.So we divided by 2.

Thus total 3 partitions possible = 10 + 15 = 25

So total number of ways possible = 25*3! = 150

2 votes
2 votes
the number of ways i to find the Number of onto functions such . the function a->b is onto if every element of b is mapped to every element of a . so here i will assume b to be 3 and a to be 5. now apply formula .

3^5-3C1*2^5 + 3c2 (1)^5.
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