how to catch duplicate random number - actionscript-3

Question

I want to display random number in bb1,bb2,bb3,bb4.
alphaset:Array = [1,2,3,4,5,6,7,8,9,10];
bb1 = randomNumber(0, alphaset.length);
bb2 = randomNumber(0, alphaset.length);
bb3 = randomNumber(0, alphaset.length);
bb4 = randomNumber(0, alphaset.length);
in above, i want to delete duplicate entry of each variable (i.e., bb1,bb2,bb3,bb4)
finally i want random number without duplicate entry of each variable
please anyone help me to achieve it.
The above four var are options for addition program. And i have to compare the above four var and bbb = randomNumber(1,10) and display bbb value in that four var randomly i,e., [bbb=bb1 or bbb=bb2 or bbb=bb3 or bbb=bb4] coz, i have to select the bbb value to display output.

Answer 1

If all the values in the array are unique to begin with, just remove them when they're used:
var list:Array = [1,2,3,4,5,6,7,8,9,10];
function getRandom(list:Array):*
{
var i:int = Math.random() * list.length;
return list.splice(i, 1)[0] || null;
}
var bb1:int = getRandom(list);
var bb2:int = getRandom(list);
var bb3:int = getRandom(list);
var bb4:int = getRandom(list);

Answer 2

One option is to shuffle the entire list of random elements - then you can take the first N elements, where N is the number you need, and know those will all be unique and randomly chosen. If you use a large amount of the total random element list, this is faster than sampling and removing.
If you can't find a standard implementation of shuffle, implement the Fisher-Yates Shuffle: http://en.wikipedia.org/wiki/Fisher%E2%80%93Yates_shuffle#The_modern_algorithm It is important that there are n! possible ways for the shuffle to come out, not n^n which you incorrectly get if you take each element one at a time and allow it to be exchanged with any other element, including previous elements, otherwise certain configurations will be more common than others.