这是一个数学问题嘛

a 每提升一个位置，就经历了 b 和 c 的全部组合，也就是 12 个
b 每提升一个位置，经历了 c 的每一项，也就是 4

所以 [2,2,4] 就是
12([1,x,x]) + 4([2,1,x]) + 4[2,2,1-4] = 20

下面写个程序验证一下

javascriptvar a = [1,2,3];
var b = [1,2,3];
var c = [1,2,3,4];

var n = 1;
for (var i = 0; i < a.length; i++) {
    for (var j = 0; j < b.length; j++) {
        for (var k = 0; k < c.length; k++) {
            console.log((n++) + ": [" + a[i] + "," + b[j] + "," + c[k] + "]");
        }
    }
}

结果（为了缩短调试，只贴了有用的部分）

plain1: [1,1,1]
...
7: [1,2,3]
...
19: [2,2,3]
20: [2,2,4]
21: [2,3,1]
...
28: [3,1,4]
29: [3,2,1]
30: [3,2,2]
...
35: [3,3,3]
36: [3,3,4]

下面再给个算位置的代码

没写注释，自己试着理解下，不明白的发评论

javascriptvar a = [1,2,3];
var b = [1,2,3];
var c = [1,2,3,4];

var all = [a,b,c];
var allCount = [b.length * c.length, c.length, 0];

function getIndex(combo) {
    var count = 0;
    for (var i = 0; i < combo.length; i++) {
        var pos = all[i].indexOf(combo[i]);
        if (pos < 0) {
            throw new Error(combo[i] + "不是有效的元素");
        }

        count += allCount[i] * pos;
    }
    count += combo[combo.length - 1];
    return count;
}

console.log("index of [2,2,4] is " + getIndex([2,2,4]));
console.log("index of [1,1,1] is " + getIndex([1,1,1]));
console.log("index of [3,3,4] is " + getIndex([3,3,4]));
console.log("index of [1,2,3] is " + getIndex([1,2,3]));
console.log("index of [3,2,1] is " + getIndex([3,2,1]));

输出（跟上面的验证输出验证下）

index of [2,2,4] is 20
index of [1,1,1] is 1
index of [3,3,4] is 36
index of [1,2,3] is 7
index of [3,2,1] is 29