Question

I was hoping someone with better math capabilities would assist me in figuring out the total possibilities for a string given it’s length and character set.

i.e. [a-f0-9]{6}

What are the possibilities for this pattern of random characters?

Answer 1

It is equal to the number of characters in the set raised to 6th power.
In Python (3.x) interpreter:

>>> len("0123456789abcdef")
16
>>> 16**6
16777216
>>>

EDIT 1:
Why 16.7 million? Well, 000000 … 999999 = 10^6 = 1M, 16/10 = 1.6 and

>>> 1.6**6
16.77721600000000 

* EDIT 2:*
To create a list in Python, do: print(['{0:06x}'.format(i) for i in range(16**6)])
However, this is too huge. Here is a simpler, shorter example:

>>> ['{0:06x}'.format(i) for i in range(100)]
['000000', '000001', '000002', '000003', '000004', '000005', '000006', '000007', '000008', '000009', '00000a', '00000b', '00000c', '00000d', '00000e', '00000f', '000010', '000011', '000012', '000013', '000014', '000015', '000016', '000017', '000018', '000019', '00001a', '00001b', '00001c', '00001d', '00001e', '00001f', '000020', '000021', '000022', '000023', '000024', '000025', '000026', '000027', '000028', '000029', '00002a', '00002b', '00002c', '00002d', '00002e', '00002f', '000030', '000031', '000032', '000033', '000034', '000035', '000036', '000037', '000038', '000039', '00003a', '00003b', '00003c', '00003d', '00003e', '00003f', '000040', '000041', '000042', '000043', '000044', '000045', '000046', '000047', '000048', '000049', '00004a', '00004b', '00004c', '00004d', '00004e', '00004f', '000050', '000051', '000052', '000053', '000054', '000055', '000056', '000057', '000058', '000059', '00005a', '00005b', '00005c', '00005d', '00005e', '00005f', '000060', '000061', '000062', '000063']
>>> 

EDIT 3:
As a function:

def generateAllHex(numDigits):
    assert(numDigits > 0)
    ceiling = 16**numDigits
    for i in range(ceiling):
        formatStr = '{0:0' + str(numDigits) + 'x}'
        print(formatStr.format(i))

This will take a while to print at numDigits = 6.
I recommend dumping this to file instead like so:

def generateAllHex(numDigits, fileName):
    assert(numDigits > 0)
    ceiling = 16**numDigits
    with open(fileName, 'w') as fout:
        for i in range(ceiling):
            formatStr = '{0:0' + str(numDigits) + 'x}'
            fout.write(formatStr.format(i))