If you only want to allow specific characters, you could also do it like this:

function randomString(length, chars) {
    var result = '';
    for (var i = length; i > 0; --i) result += chars[Math.floor(Math.random() * chars.length)];
    return result;
}
var rString = randomString(32, '0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ');

Here's a jsfiddle to demonstrate: http://jsfiddle.net/wSQBx/

Another way to do it could be to use a special string that tells the function what types of characters to use. You could do that like this:

function randomString(length, chars) {
    var mask = '';
    if (chars.indexOf('a') > -1) mask += 'abcdefghijklmnopqrstuvwxyz';
    if (chars.indexOf('A') > -1) mask += 'ABCDEFGHIJKLMNOPQRSTUVWXYZ';
    if (chars.indexOf('#') > -1) mask += '0123456789';
    if (chars.indexOf('!') > -1) mask += '~`!@#$%^&*()_+-={}[]:";\'<>?,./|\\';
    var result = '';
    for (var i = length; i > 0; --i) result += mask[Math.floor(Math.random() * mask.length)];
    return result;
}

console.log(randomString(16, 'aA'));
console.log(randomString(32, '#aA'));
console.log(randomString(64, '#A!'));

Fiddle: http://jsfiddle.net/wSQBx/2/

Alternatively, to use the base36 method as described below you could do something like this:

function randomString(length) {
    return Math.round((Math.pow(36, length + 1) - Math.random() * Math.pow(36, length))).toString(36).slice(1);
}