20 stallions, 0 mares and 80 ponies.

17 stallion, 5 mares and 78 ponies.

Stallion 14, mare 10 and pony 76.

Stallion 1 1, mare 15, pony 74.

8 stallions, 20 mares and 72 ponies.

Five stallions, 25 mares and 70 ponies.

2 stallions, 30 mares and 68 ponies.

The reasoning process is as follows: (The method may be irregular)

Suppose the stallion is X, the mare is Y and the pony is Z.

Two equations are obtained: 3X+2Y+Z/2= 100 (1).

X+Y+Z= 100 (2)

From (2): Z= 100-X-Y (3)

Substituting (3) into (2) gives 3x+2y+(100-x-y)/2 =100 (4).

Simplify equation (4): 5X+3Y= 100.

Because horses are whole, x and y must both be integers.

It can be concluded that 3Y must be a multiple of 5, and only 15, 30, 45, 60, 75, 90 can have a common multiple of 3 and 5 below 100.

Suppose 3Y= 15, then Y=5, X= 17, then Z=78.

Suppose 3Y=30, then Y= 10, X= 14, then Z=76.

Suppose 3Y=45, then Y= 15, X= 1 1, then Z=74.

Suppose 3Y=60, then Y=20, X=8, then Z=72.

Suppose 3Y=75, then Y=25, X=5, then Z=70.

Suppose 3Y=90, then Y=30, X=2, then Z=68.

Another possibility is that 100 can be divisible by 5X when 3Y=0.

At this time, Y=0 and X=20, so Z=80.