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An entrepreneur needs to assign 5 different tasks to 3 of his employees. If every employee is assigned at least 1 task, how many ways can entrepreneur assign these tasks ?


I solved it like this =>

A). Making partitions

(3,1,1) = C(5,3) * C(2,1) * C(1,1) = 20

(1,3,1) = C(5,1) * C(4,3) * C(1,1) = 20

(1,1,3) = C(5,1) * C(4,1) * C(3,3) = 20

B). Making partitions

(2,2,1) = C(5,2) * C(3,2) * C(1,1) = 30

(1,2,2) = C(5,1) * C(4,2) * C(2,2) = 30

(2,1,2) = C(5,2) * C(3,1) * C(2,2) = 30

Then total number of ways = 60 * 90 = 5400 ways


But the answer is given = 60 + 90 = 150 ways


They made addition here, but i am mainly confused when to know that we have to add or multiply. Please provide any basic idea so that one can tell that whether addition will be there or multiplication ?

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  • $\begingroup$ You shouldn't have multiplied 60 and 90 in the first place. You can't assign two ways at once! Rather, since the two ways are mutually exclusive, the total adds and you are left with the right answer. $\endgroup$ – Parcly Taxel Nov 4 '16 at 4:28
  • $\begingroup$ @ParclyTaxel, how to know multiply or addition ? $\endgroup$ – Jon Garrick Nov 4 '16 at 4:32
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Ignoring persons, the tasks can be divided into just $2$ patterns: $3-1-1\;and\; 2-2-1$

The first pattern can be chosen in $\binom53\binom21 = 20$ ways,
and allotted to individuals in $\frac{5!}{3!1!1!}$ ways to yield $60$ distributions

Similarly, the second pattern of $2-2-1$ can be distributed in $\binom52\binom32\frac{5!}{2!2!1!} = 90$ ways

We can distribute according to one or other of the two patterns,
and note that for OR cases, we always add.

You can get more guidance for when to add and when to multiply here

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