How to find subsets from a set that product equals the target?

Question

Let us say we have a list and target of:
list: [1,2,3,4,5] and target: 20 and we want to find total combinations to reach this, with multiplication, which is: [1,4,5], [4,5], [2,5,2], [1,2,2,5] I did this code, but I can't seem to know how to remove the ones so I have them too, I mean that I receive: [1,4,5], [1,2,2,5].
But without [1], I can't seem to get it, I tried to "cheat" somehow to get it, but I can't since my code doesn't fit for it...

def Coin(target, lst, temp=[], comb=[]):
    if target == 1:    
        comb.append(temp)
        return comb
    if len(lst) == 0:
        return []
    if target >= lst[0]:
        if lst[0] > 1:
            take=Coin(target/lst[0],lst,temp+[lst[0]],comb)
            dont_take=Coin(target,lst[1:],temp,comb)
        else:
            take_1 = Coin(target, lst[1:], temp + [1], comb)
            return take_1
        return take
    return comb

print(Coin(20, [1,2,3,4,5], [], []))
[[1, 2, 2, 5], [1, 4, 5]]

How to add the parts without 1? I don't need a solution, since as said, not homework, but practice for exam. Just a clue will be enough, I want to find it myself, but I need a clue for it.

Answer 1

Edit: The rules for the question is:

1. Positive integers only ( 0 not allowed ).

2. Number can appear once only ( at input ).

3. can not repeat numbers on list. EG: [1,2,3,4], n=12 >> [1,12], [12,1], [3,4], [2,6], [1,3,4], [1,2,6].
   NOT: [1,2,2,3], [2,2,3]

4. That is all I guess.

   def coin_change_MULTI(num, lst):
       if num == 1 and 1 not in lst:
           return []
       return Coin_MULTI(num, sorted(lst), [], [])
   def Coin_MULTI(target, lst, temp=[], comb=[]):
       if target == 1:
           if big_than_target(1, lst):
               return [[1]]
           comb.append(temp)
           return comb
       if len(lst) == 0: return []
       if target >= lst[0]:
           if lst[0] > 1:
               take=Coin_MULTI(target/lst[0],lst[1:],temp+[lst[0]],comb)
               dont_take=Coin_MULTI(target,lst[1:],temp,comb)
               return comb
           else:
               take_1 = Coin_MULTI(target, lst[1:], temp + [1], comb)
               dont_take_1 = Coin_MULTI(target, lst[1:], temp, comb)
               return comb
           return take + dont_take
       return comb
   
   
   
   print(coin_change_MULTI(12, [2,4,6,12,7,3, 1]))
   print(coin_change_MULTI(1, [2,4,6,12,7,3,1]))
   print(coin_change_MULTI(1, [2,4,6,12,7,3]))
   print(coin_change_MULTI(100, [2,4,6,12,7,3,1]))
   print(coin_change_MULTI(576, [2,4,6,12,7,3,1]))
   print(coin_change_MULTI(12096, [2,4,6,12,7,3,1]))
   print(coin_change_MULTI(0, [2,4,6,12,7,3,1]))
   print((coin_change_MULTI(24, [2,4,6,12,7,3,1])))
   
   
   [[1, 2, 6], [1, 3, 4], [1, 12], [2, 6], [3, 4], [12]]
   [[1]]
   []
   []
   [[1, 2, 4, 6, 12], [2, 4, 6, 12]]
   [[1, 2, 3, 4, 6, 7, 12], [2, 3, 4, 6, 7, 12]]
   []
   [[1, 2, 3, 4], [1, 2, 12], [1, 4, 6], [2, 3, 4], [2, 12], [4, 6]]
   

Process finished with exit code 0