Answer:
[tex]\frac{60!}{20!}[/tex] ways
Step-by-step explanation:
Each row of the barn can hold at most 10 rows of pens and each row can contain at most 6 pens.
A pen can contain exactly one animal.
So, each barn can hold at most 1 × 6 × 10 = 60 animals.
Available animals in the county fair = 22 cows + 18 horses = 40
40 animals can be arranged in 60 places in [tex]60P_{40}[/tex]
= [tex]\frac{60!}{(60-40)!}[/tex]
= [tex]\frac{60!}{20!}[/tex] ways