I have part of a solution. At least, it shows a way to attack the problem.
With parts 1,5,10,25,50, we have:
1 / (1-x)(1-x^5)(1-x^10)(1-x^25)(1-x^50)
=
(1+x+x^2+x^3+..)(1+x^5+x^10+x^15+..)(1+x^10+x^20+x^30+..)(1+x^25+x^50+x^75+..)(1+x^50+x^100+x^150+..)
Now, just to shorten the notation, let's consider parts a,b only.
It would generalise in the obvious way.
With parts a,b, we have:
1 / (1-x^a) (1-x^b)
=
(1+x^a+x^2a+x^3a+..) (1+x^b+x^2b+x^3b+..)
Here, the term x^(2a+3b) correspond with using a,a,b,b,b coins (summing to 2a+3b).
The exponent of x here being the sum of the parts.
So when we expand the gf (generating function) in x,
the coefficient of x^s is the number of ways to get a sum of s.
=====
Now we need to count those parts, getting the number 5.
To do that , it is best to use another variable, y.
So instead of (1 + x^a + x^2a + x^3a + ..),
we can use (1 + y x^a+ y^2 x^2a + y^3 x^3a + ..)
So with parts a,b, we have:
1 / (1 - y x^a) (1 - y x^b)
=
(1 + y x^a y + y^2 x^2a + y^3 x^3a + ..) (1 + y x^a y + y^2 x^2a + y^3 x^3a + ..)
Now, the term y^(2+3) x^(2a+3b) correspond with using a,a,b,b,b coins (counting to 2+3, summing to 2a+3b)
The exponent of y here being the number of parts.
The exponent of x here being the sum of the parts.
So when we expand the gf (generating function) in y and x,
the coefficient of y^c x^s is the number of ways to get a count of c and a sum of s .
=====
but we want an even number of coins, so we only need to count the parts modulo 2.
Instead of counting 0,1,2,3,.. we have to count 0,1,0,1,..
So instead of (1 + y x^a+ y^2 x^2a + y^3 x^3a + ..),
we can use (1 + y x^a+ x^2a + y x^3a + ..)
or (1 + x^2a + x^4a + ..) + (y x^a+ y x^3a + y x^5a ..)
or (1 + x^2a + x^4a + ..) + y (x^a+ x^3a + x^5a ..)
or (1 + x^2a + x^4a + ..) + y x^a (1+ x^2a + x^4a ..)
or (1 + y x^a) (1+ x^2a + x^4a ..)
or (1 + y x^a) / (1-x^2a)
So with parts a,b, we have:
(1 + y x^a) (1 + y x^b) / (1-x^2a) (1-x^2b)
=
(1 + y x^a y + x^2a + y x^3a + ..) (1 + y x^a y + x^2a + y x^3a + ..)
