From http://linuxsig.org/files/bash_scripting.html :
$# Stores the number of command-line arguments that
were passed to the shell program.
$? Stores the exit value of the last command that was
executed.
$0 Stores the first word of the entered command (the
name of the shell program).
$* Stores all the arguments that were entered on the
command line ($1 $2 ...).
"$@" Stores all the arguments that were entered
on the command line, individually quoted ("$1" "$2" ...).
So basically, $# is number of arguments given when your script was executed. $* is long string containing all arguments. For example $1 is first argument and so on (if you have to access specific argument in your script).
As Brian commented, there is simple example. If you run following command:
./command -yes -no /home/username
$# should be 3. $* would be "-yes -no /home/username" and $@ is an array containing
{"-yes", "-no", "/home/username"}.
As Dennis noted in comments, this assumes any bash variant as your shell.
