Cows and Chickens

Let the numbers of cows and chickens be $a$ and $b$ respectively, where $a, b \ge 1$ . We know that

$\begin{aligned} 4a+2b & = 24 \\ 2a+b &=12 \\ \implies 12-2a & = b \\ 12-2a & \ge 1 \\ 2a & \le 11 \\ \implies a & \le 5 & \small \color{#3D99F6} \text{Since }a \text{ is a natural number.} \end{aligned}$

Therefore, there are $\boxed{5}$ combinations of $(a,b) = (1,10), (2,8), (3,6), (4,4), (5,2)$ .

If there are $x$ chickens and $y$ cows, then $2x+4y=24$ , so $x+2y=12$ .

Since $2y$ is even, $x$ is even too. $x$ can't be $0$ or $12$ , so the possible values are: $2, 4, 6, 8, 10$ . This is $5$ numbers.

$24$ total, with at least one cow and one chicken, that's $6$ legs, will leave $18$ legs.

From this we could have no cows, one, two, three, or maximum four cows (which would come to $16$ legs).

That is $\boxed5$ possible options.

You can split 24 into 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 (twelve 2s).

This doesn't count as a combination, but you can take away two 2s and replace them with a 4 to get 10(2)+4=24.

You can keep doing this and replace each pair of two with a 4. So there would be 12/2 combinations = 6, minus the one combination with only 4s, to give you 5 possible combinations.