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I have adapted this question from this question, I don't fully understand the answers given there/ Understanding distribution using Stirling number of the second kind inclusion exclusion approach

Problem
I want to distribute $n=3$ distinguishable items to $r=3$ distinguishable groups such that

  • repetition of items is not allowed
  • each group contains one or more items.

Solution

We follow inclusion-exclusion principle which, in turn, takes form of Stirling number of second kind.

  • $3^3=27$ ways to distribute $3$ items to $3$ groups such that $3$ groups can contain zero or more items.

  • (Subtract) $2^3\times\binom{3}{1}=24$ ways to distribute $3$ items to $2$ groups such that $2$ groups can contain zero or more items (i.e., $1$ group is guaranteed to be empty).

  • (Add) $1^3\times \binom{3}{2}=3$ ways to distribute $3$ items to $1$ group such that $1$ group can contain zero or more items (i.e., $2$ groups are guaranteed to be empty).

Final answer $=27-24+3=6$. These six distributions corresponds to following distributions:

| Group 1 | Group 2 | Group 3 |
-------------------------------
| 1       | 2       | 3       |
| 1       | 3       | 2       |
| 2       | 1       | 3       |
| 2       | 3       | 1       |
| 3       | 1       | 2       |
| 3       | 2       | 1       |

Q1.

Does the table below correctly list all 27 possibilities?

|    | Group 1 | Group 2 | Group 3 |
|----|---------|---------|---------|
| 1  | 123     |         |         | 123 in one group
| 2  |         | 123     |         |
| 3  |         |         | 123     |
------------------------------------
| 4  | 12      | 3       |         | 12 in one group
| 5  | 12      |         | 3       |
| 6  | 3       | 12      |         |
| 7  |         | 12      | 3       |
| 8  |         | 3       | 12      |
| 9  | 3       |         | 12      |
------------------------------------
| 10 | 23      | 1       |         | 23 in one group
| 11 | 23      |         | 1       |
| 12 | 1       | 23      |         |
| 13 |         | 23      | 1       |
| 14 |         | 1       | 23      |
| 15 | 1       |         | 23      |
------------------------------------
| 16 | 13      | 2       |         | 13 in one group
| 17 | 13      |         | 2       |
| 18 | 2       | 13      |         |
| 19 |         | 13      | 2       |
| 20 |         | 2       | 13      |
| 21 | 2       |         | 13      |
------------------------------------
| 22 | 1       | 2       | 3       | All digits/items in different groups
| 23 | 1       | 3       | 2       |
| 24 | 2       | 1       | 3       |
| 25 | 2       | 3       | 1       |
| 26 | 3       | 1       | 2       |
| 27 | 3       | 2       | 1       |

Q2.

How so these lines to the (24 &3) correspond to?

  • Does the 3 correspond to the first three lines?

  • Does the 24 correspond to lines 4 to 27?

Q3. Where does that $24$ come from?

  • The first part is because there are 3 ways to exclude one bin = 3
  • I am struggling to understand the groups of eight though can someone please explain how they are formed?
RobPratt
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Bazman
  • 899

1 Answers1

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Q1: Looks OK to me.

Q2: The 24 corresponds to lines 1..21, with lines 1..3 counted twice because each of those can occur two ways, depending on which group we "guarantee to be empty". The 3 corresponds to lines 1..3.

Q3: We can pick one bin to guarantee the emptiness of, in $\binom{3}{1}$ ways. Then we can pick how to allocate the 3 objects to the 2 remaining bins in $2^3$ ways. The thing that I think you're finding strange is that this counts some possibilities twice, namely those where more than one bin is empty.

This sort of "overcounting" is always there in inclusion-exclusion situations. You overcount, then you correct, then you find you overcorrected so you correct for that, etc., and the inclusion-exclusion principle is a way of doing the bookkeeping to make sure that you end up counting everything exactly once.

Gareth McCaughan
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