In bash, how can I print the first n elements of a list?

Question

In bash, how can I print the first n elements of a list?

For example, the first 10 files in this list:

FILES=$(ls)

Answer 1

My way would be:

ls | head -10 | tr "\n" " "

This will print the first 10 lines returned by ls, and then tr replaces all line breaks with spaces. Output will be on a single line.