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