The book i'm reading says the following :
Number of ways in which $(m+n+p)$ items can be distributed into $m$, $n$ and $p$ sized groups is $$\frac{(m+n+p)!}{m!n!p!}$$ I understand this is called the book-keeper's rule.
So, following from this, if m=n=p, then this formula should become $$\frac{(3n)!}{(n!)^3}$$
Generalising, if we have m groups of n items each, the formula should be $$\frac{(mn)!}{(n!)^m}$$ BUT the book breaks into a new section, and says :
The number of ways $mn$ stuff can be divided equally into $m$ groups of $n$ item each is :
$$\frac{(mn)!}{(n!)^m(m!)}$$
Whereas, The number of ways $mn$ stuff can be distributed equally into $m$ groups of $n$ item each is :
$$\frac{(mn)!}{(n!)^m}$$
Why the two different formulas ? Can someone give a logical example/explanation of the difference here ? Between divided and distributed ?
ps : my LaTex sucks. Sorry about the horrible look of the post.
Best Answer
If $n\neq m\neq p$, then the total amount of different orderings is $(m+n+p)!$. However, in this case we get the same group of $m$ elements $m!$ times. The same for $n$ and $p$ and thus we have $(m+n+p)!/(m!n!p!)$ amount of distinct orderings into groups. Suppose now however that $m=n=p$. Then we not only need to take into consideration that we get the same group $n!$ times for a certain $n$, but also that we can interchange our three groups, since they are the same size. Since we have three groups, we have $3!$ types of orders and thus we have $(3n)!/((n!)^33!)$.
Example:
Suppose we look in how many ways we can divide four elements in $3,1$ groups: The total number of orderings is $4!$, but $123,4$ equals $132,4$ and thus we get the following distinct groups \begin{align} 123,4 \\ 124,3 \\ 134,2 \\ 234,1 \end{align} for a total of $4!/(3!1!) = 4$. However, suppose now we want to divide in 2,2 groups. Then not only is $12,34$ equal to $21,34$. It is also equal to $34,12$ ie switching the groups. Therefore we get \begin{align} 12,34 \\ 13,24 \\ 14,23 \end{align} for a total of $4!/((2!)^22!) = 3$.