Question: I'm trying to find all containers that does not have any resources spec in the cluster and make an alert for that.  

Now I'm thinking about use all `container` and `pod` label in `kube_pod_container_info` query and do an `absent` query for `kube_pod_container_resource_requests`,but dont know how to achieve that.

Answer: 
You are allowing repeats (for example, [1,1,1] and [2,2,2]). The value of 60 is for permutations without repeats. You do that by checking that you aren't repeating a value.

**NOTE** that this code only works if there are no repeats in `x`. If there are duplicates, then you would have to use indexes instead (that is, `for i in range(len(x)):`).

```
x = [1,2,3,4,5]
y = []
for i in x:
    for j in x:
        if i == j:
            continue
        for k in x:
            if i!=k and j!= k: 
                y.append((i,j,k))
print(y)

```
As pointed out by Tim Roberts, I was adding repeats of `i,j or k`, `(1,1,1)`. My fix is to just make sure `i,j and k` are different:

```py
for i in x:
    for j in x:
        for k in x:
            # If i,j,k are different
            if len(set((i, j, k))) == 3:
                y.append((i, j, k))

```

As `set((i, j, k))` will remove the duplicates in the tuple `(i, j, k)`, so the length must equal 3. This is also helpful if I need to make this recursive for `nPr(n,r)` as `set((i, j, k ... r)) == r`.