Generate a dictionary of passwords

I have a Python script that generates a dictionary of all possible 3 digit password combinations. It essentially works by just nesting for loops. This is fine, but say I wanted to change it so it could generate 4, 5, 6, ... digit combinations. This would require me to keep adding more nested loops and changing the print statement.

How can I make my current code more efficient and apply the DRY principle? This way, if I wanted to change it to generate 4-digit codes, I could simply change a single variable rather then add an entire for loop.

Current code:

f=open('New Text Document.txt','w')
def getCombos(digits, caps, numbers):
    if(caps == True):
        if(numbers == True):
            for x in range(33, 126):
               for y in range(33, 126):
                   for z in range(33, 126):
                       print(chr(x) + "" + chr(y) + "" + chr(z))
                       f.write(chr(x) + "" + chr(y) + "" + chr(z) + "\n")

getCombos(3, True, True)

If I wanted to add a fourth digit:

f=open('New Text Document.txt','w')
def getCombos(digits, caps, numbers):
    if(caps == True):
        if(numbers == True):
            for x in range(33, 126):
               for y in range(33, 126):
                   for z in range(33, 126):
                       for m in range(33, 126):
                       print(chr(x) + "" + chr(y) + "" + chr(z) + "" + char(m))

getCombos(3, True, True)

Please take a look at itertools.product. It will simplify your code greatly.

def getCombos(digits, caps, numbers):
    # Build set of choices appropriately based on caps and numbers
    choices = list(range(33, 126))
    for vals in itertools.product(choices, repeat=digits):
        print ''.join(map(chr, vals))

getCombos(3, True, True)

StackExchange Code Review Q#119327, answer score: 15