How many combinations of numbers and letters 5-20 characters are there?

Question

ex:
abcde
abcdf
abcdef
abcd1
12345
abcde12345fghij67890

etc... how many are there?

How did you find out?

Yes you can repeat characters

Answer 1

You are looking for the combinations for "words" formed byt letters or numbers with 5, 6, 7... 20 characters, right?
So, the combinations would be the sum of the words of 5 characters plus the ones with 6 plus the one with 7 and so so until 20
For the ones with 5 characters you have 5 spaces, __ __ __ __ __ in each one you have 36 possibilities (26 letters plus numbers 0-9). For that reason, the result for this is 36^5 = 60466176
Note: With this operation we are supposing that 0 could be first character too, if not would be 35 in the first space and 36 in the others, so 35 x 36^4
For 6 characters the answer would be 36^6
For 7 would be 36^7
and so so...
note that if 0 cant be a first character you would have to do the similar thing I explained you before
The final answer would be:
36^5 + 36^6 + 36^7 + 36^8 + 36^9 + 36^10 + 36^11 + 36^12 + 36^13 + 36^14.... + ^36^20

If 0 can not be a first character, the final answer would be:
35 x (36^4 + 36^5 + 36^6 + 36^7+... + 36^19) because the first one just have 35 possibilities

Any doubt?

Answer 2

Since you said you can repeat characters, the answer is NEITHER a permutation NOR a combination.

There are 26 letters and 10 digits, so for every position, you have 36 choices.

Therefore the answer is:

36^5 + 36^6 + 36^7 + 36^8 + 36^9 + 36^10 + 36^11 + 36^12 + 36^13 + 36^14 + 36^15 + 36^16 + 36^17 + 36^18 + 36^19 + 36^20

According to Google, that number is about 1.3749423 × 10^31.

Answer 3

I think I know what you want, but I'm leaving
repeating characters out as it's too involved.

There are 26 letters and 10 digits
for a total of 36 characters.

Combinations give no attention to the order of arrangement.

So choosing 2 characters from abcd gives
ab, ac, ad, bc, bd and cd,
that is, only 6 combinations.

Permutations give attention to the order of arrangement.

So choosing 2 characters from abcd gives
ab, ba, ac, ca, ad, da, bc, cb, bd, db, cd and dc,
that is, 12 permutations.

The formulae for each of these are :

Combinations : nCr = n! / [r! * (n - r)!]

Permutations : nPr = n! / (n - r)!

This means the number of combinations (or permutations)
of n different objects taken r at a time.
! is the factorial sign.

Here are the results I think you want :
You have to choose whether you
want combinations or permutations.

36C5 = 376992
36C6 = 1947792
36C7 = 8347680
36C8 = 30260340
36C9 = 94143280
36C10 = 254186856
36C11 = 600805296
36C12 = 1251677700
36C13 = 2310789600
36C14 = 3796297200
36C15 = 5567902560
36C16 = 7307872110
36C17 = 8597496600
36C18 = 9075135300
36C19 = 8597496600
36C20 = 7307872110

36P5 = 45239040
36P6 = 1402410240
36P7 = 42072307200
36P8 = 1220096908800
36P9 = 34162713446400
36P10 = 922393263052800
36P11 = 23982224839372800
36P12 = 599555620984320000
36P13 = 14389334903623680000
36P14 = 330954702783344640000
36P15 = 7281003461233582080000
36P16 = 152901072685905223680000
36P17 = 3058021453718104473600000
36P18 = 58102407620643984998400000
36P19 = 1045843337171591729971200000
36P20 = 17779336731917059409510400000

Answer 4

Do 'a' and 'A' count as 1 or 2?

There are (52 + 10) options per position
62^5 + 62^6... 62^20 = 7.15971398 × 10^35

If it's 1 then the solution has already been given in a post above.

Answer 5

Are you repeating characters? (i.e. does aaaaa count)

You'll either be using permutations or combinations depending if repeated charaters count. I'll just do the permutations.

So, you have 26 letters and 10 numbers to choose from, 36 characters total. The first set you're taking 5 of those, then 6... up to 20.

36P5 + 36P6 + 36P7 + ... + 36P20 = the total number of permutations of 5-20 character strings choosen from numeric and alpha characters.

Answer 6

about.... hmmm lets say..... 3786741241786... thats waht i got... i used my calculator.