# Confusion in formula for division into groups

My book says:

The number of ways of distributing $m+n+p$ different things among three persons in unequal groups containing m,n, and p things is:

$$\frac{(m+n+p)!}{m!*n!*p!}*3!$$

which makes sense to me because it's equivalent to product of:

1. choosing one person and giving him m objects in $^3C_1*^{m+n+p}C_m$ ways
2. then choosing second person and giving him n objects in $^2C_1*^{n+p}C_n$ ways
3. then choosing third person and giving him p objects in $^1C_1*^{p}C_p$ ways

But then the book also gives another formula saying:

The number of ways of distributing $3m$ different things among three persons in equal groups containing m things is $$\frac{(3m)!}{(m!)^3}$$

Notice that it's missing $*3!$. According to me it should be: $\frac{(3m)!}{(m!)^3}*3!$, but it is not. That means it does not follow from the previous generalized formula ($\frac{(m+n+p)!}{m!*n!*p!}*3!$). My thinking says that it should have the $*3!$ as, the above formula is equivalent to product of:

1. choosing one person and giving him m objects in $^3C_1*^{3m}C_m$ ways
2. then choosing second person and giving him m objects in $^2C_1*^{2m}C_m$ ways
3. then choosing third person and giving him m objects in $^1C_1*^{m}C_m$ ways

My question is:

Why does the second formula not have $*3!$ in it? Where did my thinking go wrong?

If there are $3m$ different things to be distributed among $3$ persons and each of them gets $m$ things then there are $\frac{(3m)!}{(m!)^3}$ ways to do that. Let's think of persons $1,2,3$. First $m$ things are selected for person $1$. There are $\binom{3m}{m}$ possibilities. Then from the remaining things $m$ are selected for person $2$. There are $\binom{2m}m$ possibilities. The rest is for person $3$ and we come to a total of: $$\binom{3m}{m}\binom{2m}{m}=\frac{(3m)!}{(m!)^3}$$ possibilities.
If we split up $m+n+p$ things among $1,2,3$ such that one of them gets $m$ one of them gets $n$ and one of them gets $p$ things then we can start with going through the same process in the sense that first $m$ things are selected of person $1$, then $n$ for person $2$ and the remaining things are for person $3$. This leads to: $\binom{m+n+p}{m}\binom{n+p}{n}=\frac{(m+n+p)!} {m!n!p!}$ possibilities. However we only counted the splitups of the form $m$ for $1$, $n$ for $2$ and $p$ for $3$. If $m,n,p$ are distinct then we need an extra factor $3!$ to repair that and we end up with:$$\frac{(m+n+p)!3!} {m!n!p!}$$ possibilities.
• If there are persons 1,2,3 then there are $\frac{(m+n+p)!}{m!n!p!}$ distributions s.t. person 1 gets $m$, person 2 gets $n$ and person 3 gets $p$ things (first round). But also there the distributions that person 1 get $m$ person 2 gets $p$ and person 3 gets $n$ things (second round). In fact $3!=6$ rounds in total. However, if $m=n=p$ then we are ready after the first round. Try it out on $m=n=p=1$ to get some understanding of it. There $6$ is definitely the answer. Not $36$. I admit that my answer is not very clear in this. Later I will try to make it better. – drhab Aug 9 '16 at 13:55