Selecting sample from a list of n numbers without repeating

So lets say I have a python list:

num_list = ["3", "2", "5", "6", "9", "3", "0"]

What I want to do is:
each time select a sample with 5 numbers from this list;
the first number can only be 0, 3, or 9.

How can I write an algorithm in python find all the combinations that can be sampled from the list with the above two conditions?

I have used the random module in python, but it can't do restricted sample selection like the condition for first number being only 0, 3 or 9.

create a for loop with elements = ['0', '3', '9']
then make a variable x = [element]
then use random.choice as usual

    num_list = ['3','2','5','6','9','3','0']
    combinations = []
    elements = ['0', '3', '9']
    for element in elements:
        x = [element]
        while len(x) < 5:
            x.append(random.choice(num_list))
        else:
            combinations.append(x)
        # you need to make the part under for loop run till all combinations
        # are stored
        # you also need to remove the current element from the list or else
        # it will be considered twice, to do this make a copy of list and 
        # remove the current element in the for loop

It isn't clear if you want all possible arrangements of that list of or truly all permutations from a random selection. That is, can you sample the same number from the list twice or are the numbers essentially being popped of the list at selection time.

Let's assume you need every combination of length 5 where 0, 3, or 9 begin the list.

from itertools import combinations

num_list = ["3", "2", "5", "6", "9", "3", "0"]

c = combinations(num_list, 5)
candidates = [l for l in c if l[0] in ["0", "3", "9"]]

This yields a list of candidates that is 15 items long.

If instead you want all possible length 5 permutations of that list.

from itertools import permutations

num_list = ["3", "2", "5", "6", "9", "3", "0"]

p = permutations(num_list, 5)
candidates = [l for l in p if l[0] in ["0", "3", "9"]]

That yields a list of canbidates that is 1440 items long as it allows for repeated elements.

So these allow you to find all possible, if you need to find all possible. But if you need to do so randomly without finding all possible, no sense wasting the memory and time on list, just pull them from the itertools objects directly as they are generators that will yield as required.

You don't need to sample from all the combinations (which would quickly become prohibitive as you increase the number of items). Instead, separate the selection of the first item from sampling of the rest. To do so, select the first number from a random choice of the indexes that contain 0, 3 or 9. Then pick a sample of 4 from the remaining positions

nums = ["3", "2", "5", "6", "9", "3", "0"]

import random

firstIndexes = [i for i,n in enumerate(nums) if n in "039"]
for _ in range(10):
   i0 = random.choice(firstIndexes)
   result = (nums[i0],*random.sample(nums[:i0]+nums[i0+1:],4))
   print(result)

('9', '2', '0', '6', '5')
('3', '0', '6', '2', '9')
('9', '5', '3', '3', '2')
('3', '0', '9', '3', '5')
('9', '0', '2', '3', '6')
('3', '6', '5', '2', '3')
('0', '9', '3', '3', '6')
('3', '2', '3', '9', '6')
('9', '3', '6', '2', '3')
('9', '5', '0', '2', '3')

One way would be to use a simple loop to iterate until you get one of the selected numbers:

number = None
while True:
    number = random.choice(arr)
    if number in [0, 3, 9]:
        break

Then use the random.choice as usual for the rest of the numbers.