Usbrth

I have a set of $n$ discrete variables $mathcalA = a_1, dots, a_n $, where every $a_i in mathcalA$ can take on the values in $ 0, 1, dots, m $ and $n,m in mathbbN$. I wish to create a joint measure -- an unnormalised probability distribution -- over all joint events described by the variables and then determine a marginal probability distribution over some $a_i in mathcalA$. Each event must obey the following rule: $a_i neq a_j$, if $a_i = k$ and $k > 0.$ That is to say, multiple variables may equal zero in any event, but every non-zero value must be unique. Here is an example for $n = 3$ and $m = 2$:
beginarrayccc hline
a_1 & a_2 & a_3 & phi(a_1, a_2, a_3) \ hline
0 & 0 & 0 & 1 \
0 & 0 & 1 & 1 \
0 & 0 & 2 & 1 \
0 & 1 & 0 & 1 \
0 & 1 & 2 & 1 \
0 & 2 & 0 & 1 \
0 & 2 & 1 & 1 \
1 & 0 & 0 & 1 \
1 & 0 & 2 & 1 \
1 & 2 & 0 & 1 \
2 & 0 & 0 & 1 \
2 & 0 & 1 & 1 \
2 & 1 & 0 & 1 \ hline
& textElsewhere & & 0\ hline
endarray
I wish to create a similar joint measure for an arbitrary $n$ and $m$ and then determine the marginal distribution, $$P(a_i) = frac1K sumlimits_mathcalA - a_i phi(mathcalA) text,$$ where $$ K = sum_mathcalA phi (mathcalA) = sum_k=0^min(m, n) k! binomnk binommk $$ is a normalising constant and $mathcalA - a_i$ denotes the set difference. Quite obviously, all marginal distributions over every $a_j in mathcalA$ are equivalent, as the joint measure is symmetrical. Is it possible to find an expression for this marginal distribution, as I have done for the normalising constant?

Edit. I believe the marginal distribution is given by $$ P(a_i = 0) = frac1K sumlimits_k=0^min(m, n-1) k! binomn-1k binommk $$ and $$ P(a_i neq 0) = frac1K sumlimits_k=0^min(m-1, n-1) k! binomn-1k binomm-1k text;$$ however, I would like to show that $$sum_j=0^m P(a_i = j) = 1 text.$$

Here's how you prove this is a probability distribution. I have removed the upper limits of the summations; this is allowed since all subsequent terms are zero.
beginalign
sum_j=0^m P(a_i=j)
&=P(a_i=0)+mP(a_ineq 0)
\&=frac1K sum_kge 0 k! binomn-1k binommk+mcdotfrac1Ksum_kge 0k!binomn-1kbinomm-1k.
endalign
Multiplying by $K$,
beginalign
Ksum_j=0^m P(a_i=j)
&= sum_kge 0 k! binomn-1k binommk+sum_kge 0k!binomn-1kcdot mcdot binomm-1k
\ &= sum_kge 0 k! binomn-1k binommk+sum_kge 0k!binomn-1kcdot (k+1)binommk+1
\ &= sum_kge 0 k! binomn-1k binommk+sum_kge 0(k+1)!binomn-1kbinommk+1
\ &= sum_kge 0 k! binomn-1k binommk+sum_kge 1k!binomn-1k-1binommk
\ &= 0!binomn-10binomm0+sum_kge 1 k! left[binomn-1k+ binomn-1k-1right]binommk
\ &= 0!binomn-10binomm0+sum_kge 1 k! binomnkbinommk
\ &= sum_kge 0 k! binomnkbinommk
\&=K.
endalign
Now divide by $K$.