128. The trainer brings three lions A, B, C and two tigers D and E onto the circus arena and seats them in a row on pedestals. At the same time, the tigers cannot be placed next to each other, otherwise a fight between them is inevitable. How many ways are there to arrange the animals?
$\triangle$ First, let's calculate the number of ways to arrange the tigers. If tiger G takes the first place, then tiger D can take the third, fourth, or fifth place. If G takes the second place, then D can only take the fourth or fifth place. If $\Gamma$ is placed on the third or fourth place, then in each of these cases, D can be placed in one of two places. If G takes the fifth place, then there are three places left for D. Therefore, the number of ways to arrange the pair of tigers is

$$
3+2+2+2+3=12
$$

Now, let's deal with the arrangement of the lions. Lion A can be placed in any of the three remaining places after the tigers are arranged; then lion B can be placed in any of the two remaining places; lion C will get the last free place. Thus, the number of ways to arrange the lions, according to the rule of product, is $3 \cdot 2 \cdot 1=6$.

Therefore, the total number of ways to arrange the animals, based on the same rule, is $12 \cdot 6=72$.

Answer: 72.