how do I execute a program which take 1 or more arguments in a shell script??

zlly20

I have a problem that I need to execute a program in my shell and pass some arguments to the program..
I have written a simple shell as follows:

#!/bin/bash

HALLPATH=/home/yzhang/

$($HALLPATH/hall_picker)

hall_picker is the program I need to execute, and this program needs at least one argument to run.
so, how do I write my shell script to pass the argument to the program when the shell script is execute in following format :

nameofthescript argument1 argement2 argement3

I tried this
$($HALLPATH/hall_picker $1 $2 $3) in the script, but it does not work....
what is the right way to do it??? please help....