Dear Fabrice,

You may split Binom(n,k) function into two functions as following:

*****
\startluacode
P={}
combi = P

function P.fact (n)
  if n <= 0 then
    return 1
  else
    return n * P.fact(n-1)
  end
end

function P.ncr(n,r)
    return P.fact(n)/(P.fact(r)*P.fact(n-r))
end
combi = {
    fact = fact,
    ncr = ncr,
}
\stopluacode
******

Your table is actually Pascal’d triangle. 
Using the above function, I was able to draw Pascal’s triangles.
Hans helped me to complete it.
I couldn’t wikify it at that time because I don’t know how to.  I’ll do it soon.

Here is the whole code for Pascal’s triangle in two different ways using Lua, Metafun and ConTeXt.
You may enhance the code.

\startbuffer[pt1]
numeric n,r,s,u,dx,dy,tt; u := 1.8cm;
path p, q;
pair A,B,start,now;
A := dir(210)*u;
B := dir(-30)*u;
dy := sind(30)*u;
dx := 2*cosd(30)*u;
for n=0 upto 4:
    start := n*dir(210)*u;
    for r=0 upto n:
        s := n-r;
        tt := lua("mp.print(P.ncr(" & decimal n & "," &  decimal r & " ))");
        now := start+r*right*dx;
        dotlabel.top(textext("$\displaystyle {" & decimal n & "\choose" & decimal r & "} = "& decimal tt & "$"),now);
        draw (now+A) -- now -- (now+B);
    endfor;
endfor;
\stopbuffer
\startbuffer[pt2]
numeric n,r,s,u,dx,dy,tt; u := 1cm;
path p, q;
pair A,B,start,now;
A := dir(210)*u;
B := dir(-30)*u;
dy := sind(30)*u;
dx := 2*cosd(30)*u;
for n=0 upto 8:
    start := n*dir(210)*u;
    for r=0 upto n:
        s := n-r;
        tt := lua("mp.print(P.ncr(" & decimal n & "," &  decimal r & " ))");        
        now := start+r*right*dx;
        label(textext("$" & decimal tt & "$"),now);
    endfor;
endfor;
\stopbuffer
\startluacode
P={}
combi = P

function P.fact (n)
  if n <= 0 then
    return 1
  else
    return n * P.fact(n-1)
  end
end

function P.ncr(n,r)
  return P.fact(n)/(P.fact(r)*P.fact(n-r))
end
combi = {
  fact = fact,
  ncr = ncr,
}
\stopluacode
\starttext

\processMPbuffer[pt1]
\blank[big]
\processMPbuffer[pt2]

\stoptext

I hope that it helps.

Best regards,

Dalyoung