To solve this problem we need to start by turning the available information into equations. Let the number of animals be c, p, h, and r for cows, pigs, chickens, and rabbits respectively. Then we have:

c + p + h + r = 100
10c + 3p + h + 0.5r = 100
c > p
h > r

Let's also assume that there is at least 1 of each animal, since otherwise it would not be on the list. From the second equation we can see that 1 <= c <= 9 (i. e., the number of cows must be between 1 and 9). In fact, since c > p and p > 0, there must be at least 2 cows. This leads to 8 cases as follows:

Case 1: c = 2

2 + p + h + r = 100
20 + 3p + h + 0.5r = 100
Therefore, –2p + 0.5r = 18
Since the number of pigs is less than the number of cows, p = 1. So r = 40, and h = 57. This solution satisfies all the conditions of the problem. But is it unique?

Case 2: c = 3

3 + p + h + r = 100
30 + 3p + h + 0.5r = 100
Therefore, –2p + 0.5r = 27
Solving for r gives r = 54 + 4p. Since p > 0, the number of rabbits must be greater than 54. But if r > 54, then there cannot be more chickens than rabbits because that would put the total number of animals over 100. Therefore, Case 2 fails to give a solution.

Cases 3–8: c > 3

Combining the first two equations where c is known gives
–2p + 0.5r = (100 – c) – (100 – 10c) = 9c
Solving for r gives r = 18c + 4p. As c increases, r increases too, so we can apply the reasoning from Case 2 to show that r is never less than 50, so h cannot be greater than r. Therefore, none of the remaining cases lead to a solution.

The solution from Case 1 is unique. The list must have called for:

2 Cows @
\$10 ea. = \$20
1 Pig @
\$3 ea. = \$3
57 Chickens @
\$1 ea. = \$57
40 Rabbits @
50¢ ea. = \$20

100 animals for \$100