1
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

edit (I washed away my initial answer wich caused confusion)

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.

$\endgroup$
  • $\begingroup$ "If the persons that receive the things are distinghuishable on base of the numbers of things" But, I was taught that persons are always distinguishable, while heaps/stacks are not. For example, we have two sets (of 5 different cards) X and Y. Giving X to person A and Y to person B is not the same as X to person B and Y to person A, even though it is the same number of cards. However, if we do this on stacks, then since stacks have no individual identity of their own, the above 2 distribution cases will be same. Please clear my confusion! $\endgroup$ – Gaurang Tandon Aug 9 '16 at 13:11
  • $\begingroup$ 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. $\endgroup$ – drhab Aug 9 '16 at 13:55
  • $\begingroup$ I have changed my answer. $\endgroup$ – drhab Aug 10 '16 at 8:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.