0
$\begingroup$

I was reading about distribution of distinguishable objects into indistinguishable groups. I have found following somewhat intuitive formulae when I went through various sources:

enter image description here

I have few doubts:

  1. Are these formulae correct? (I am able to understand them, but still need confirmation).

  2. Sheldon Ross's book says "disitribution of Distinguishable objects into indistinguishable groups" leads to Stirling's number of second form. Then how this relate to the last two rows in above table as they also talk about "dsitribution of Distinguishable objects into indistinguishable groups"? Whats the difference? Is it that the groups in above table are indistinguishable because of their same size and the groups in the case of Stirlings number are indistinguishable for no reason, and they may be of same size or of different size?

  • 0
    (1) Yes. (2) The table discusses ways of grouping objects into groups of *specified* sizes. Stirling numbers count ways of group objects into nonempty groups of any sizes. – 2017-02-04
  • 0
    So do you mean, in case of Stirling number groups may be of same size or of different, just that they should not be empty? And for all rows in above table, groups can be empty? – 2017-02-04
  • 0
    In all of the entries in your table, **the size of each group is specified beforehand**. For example, $\binom{5}{1,2,2}$ specifies the groups must have sizes $1$, $2$ and $2$. But Stirling numbers count groupings that do not have any fixed sizes for their groups. For example $S(5,3)$ will count groupings of five things into groups of sizes $1,2,2$ but it will *also* count groupings into groups of sizes $1,1,3$. – 2017-02-04
  • 0
    ohh...ohh...its about specified size vs any/unspecified size then? – 2017-02-04
  • 1
    Yes. ${}{}{}{}$ – 2017-02-04
  • 0
    What we can comment about order of distribution in case of Stirlings case? (I feel order does not matter.) – 2017-02-04
  • 1
    It does not matter. – 2017-02-04

0 Answers 0