Finding the number of solutions to a diophantine equation

There is an especially useful function for this kind of task: FrobeniusSolve[{a, b, c}, d] for finding the list of all solutions to the equation a x + b y + c z == d, where a,b,c are given positive integers and d is an integer, while x,y,z are non-negative integers to be found. There are many solutions (884 of them):

FrobeniusSolve[{2, 3, 1}, 100] // Short 

{{0, 0, 100}, {0, 1, 97}, {0, 2, 94}, {0, 3, 91}, <<876>>, {48, 0, 4}, {48, 1, 1}, {49, 0, 2}, {50, 0, 0}} 

Note that I used Short which omitted 876 solutions. If you need only positive solutions you can use e.g. DeleteCases to get lists free of 0:

DeleteCases[ FrobeniusSolve[{2, 3, 1}, 100], {___, 0, ___}] // Length 

 784 

Edit

FrobeniusSolve is not a superfluous function since it is much more efficient than an adequate use of Solve. Consider e.g. the analogous equation 2 x + 3 y + z == 1000. We compare the timings of solving the same equation:

Solve[{2*x + 3*y + z == 1000, x >= 0, y >= 0, z >= 0}, {x, y, z}, Integers] // Length // AbsoluteTiming FrobeniusSolve[ {2, 3, 1}, 1000] // Length // AbsoluteTiming 

 {76.9510000, 83834} {0.1900000, 83834} 

This huge efficiency difference doesn't really change if we are to find positive solutions:

Solve[{2*x + 3*y + z == 1000, x > 0, y > 0, z > 0}, {x, y, z}, Integers] // Length // AbsoluteTiming DeleteCases[ FrobeniusSolve[{2, 3, 1}, 1000], {___, 0, ___}] // Length // AbsoluteTiming 

 {77.9720000, 82834} {0.2420000, 82834} 

Moreover, the lists of solutions have the same orderings; e.g.:

DeleteCases[ FrobeniusSolve[{2, 3, 1}, 200], {___, 0, ___}] === Solve[{2*x + 3*y + z == 200, x > 0, y > 0, z > 0}, {x, y, z}, Integers][[All, All, 2]] 

True 

There are also power series methods for counting these.

SeriesCoefficient[ x^(1 + 2 + 3)/(1 - x^1)*1/(1 - x^2)*1/(1 - x^3), {x, 0, 100}] (* Out[118]= 784 *) 

See also "Supplement to 'Perplexities Related to Fourier's 17 Line Problem'."