There are exactly ten ways of selecting three from five, 12345:

123, 124, 125, 134, 135, 145, 234, 235, 245, and 345

In combinatorics, we use the notation, ⁵C₃ = 10.

In general,

nCr =

n! r!(nr)!

,where r n, n! = n(n1)...321, and 0! = 1.

It is not until n = 23, that a value exceeds one-million: ²³C₁₀ = 1144066.

How many, not necessarily distinct, values of  ⁿC_r, for 1 n 100, are greater than one-million?

+-%run_procs
run_procs(wait,0,_,Val)->Val;
run_procs(wait,Running,AccFun,Val)->
	NextVal=receive Anything->Anything end,
	run_procs(wait,Running-1,AccFun,AccFun(Val,NextVal));
run_procs(HowMany,ArgFun,AccumulatorFun,InitialAccVal)->
	run_procs(1,HowMany+1,ArgFun,AccumulatorFun,InitialAccVal).
run_procs(Limit,Limit,_,AccFun,Val)->run_procs(wait,Limit-1,AccFun,Val);
run_procs(I,Limit,ArgFun,AccFun,Val)->
	[Mod,Fun,Args]=ArgFun(I),
	spawn(Mod,Fun,Args),
	run_procs(I+1,Limit,ArgFun,AccFun,Val).

+-%choose
choose(N,N)->1;
choose(0,_)->1;
choose(N,R)->
	product(lists:seq(1+lists:max([R,N-R]),N)) div product(lists:seq(1,lists:min([R,N-R]))).

+-%product
product(List)->
	lists:foldl(fun(X,Prod)->X*Prod end, 1, List).

p53()->
	Ans=run_procs(100,fun(I)->[euler,p53,[I,1,0,self()]] end,
			  fun(A,{done,B})->A+B end,0),
	io:format("~w~n",[Ans]).
p53(N,N,Total,Parent)->Parent!{done,Total};
p53(N,R,Total,Parent)->
	C=choose(N,R),
	if
		C>1000000->p53(N,R+1,Total+1,Parent);
		true->p53(N,R+1,Total,Parent)
	end.

+-#choose
def choose(n,r)
	return 1 if r==0 or r==n
	a=[r,n-r].sort
	(((a[1]+1)..n).inject{|u,v|u*v})/factorial(a[0])
end

+-#factorial
def factorial(n)
  return 1 if n<2
  (2..n).inject{|u,v|u*v}
end

def p53
  res=0
  (1..100).each do |n|
    (1...n).each do |r|
      res+=1 if choose(n,r)>1000000
    end
  end
  puts res
end