Bash script - variable content as a command to run

line=$((${RANDOM} % $(wc -l < /etc/passwd)))
sed -n "${line}p" /etc/passwd

just with your file instead.

In this example I used the file /etc/password, using the special variable ${RANDOM} (about which I learned here), and the sed expression you had, only difference is that I am using double quotes instead of single to allow the variable expansion.

You're are probably looking for eval $var.

You just need to do:

#!/bin/bash
count=$(cat last_queries.txt | wc -l)
$(perl test.pl test2 $count)

However, if you want to call your Perl command later, and that's why you want to assign it to a variable, then:

#!/bin/bash
count=$(cat last_queries.txt | wc -l)
var="perl test.pl test2 $count" # You need double quotes to get your $count value substituted.

...stuff...

eval $var

As per Bash's help:

~$ help eval
eval: eval [arg ...]
    Execute arguments as a shell command.

    Combine ARGs into a single string, use the result as input to the shell,
    and execute the resulting commands.

    Exit Status:
    Returns exit status of command or success if command is null.