The number of strings of six characters (uppercase alphabets and numbers) out of which at least one character should be a number. This problem is often solved like this:

All strings of $6$ digits made up of uppercase characters and numbers $= 36^6$.

All strings of $6$ digits with only uppercase characters = $26^6$.

Answer $= 36^6 - 26^6 = 1867866560$.

I was searching for a way to do this not through the subtraction way as above.

  • $\begingroup$ I am afraid that you won't get anything simpler than this. $\endgroup$ – Nathanael Skrepek Apr 6 '17 at 9:01
  • 1
    $\begingroup$ I don't want a simpler solution, I am just asking for one $\endgroup$ – Sanjay Apr 6 '17 at 9:02
  • $\begingroup$ add up the 1,2,3,4,5,6 number cases (obviously more complicated) $\endgroup$ – Cato Apr 6 '17 at 9:15

The subtraction way is definitely the right approach. If you really wanted to do it a different way the following would work.

If the first digit occurs in position 1, there are $10\times 36^5$ options (digit followed by anything). If the first digit is in position 2, there are $26\times10\times 36^4$ options (letter, digit, then anything), and so on, so the answer is

$$10\times 36^5+26\times10\times36^4+26^2\times10\times36^3+26^3\times10\times36^2+26^4\times10\times36+26^5\times 10.$$

  • $\begingroup$ I only understand this till option 1. I mean, why can't it be like this : 10x36^5 (number at first position), 36 x 10 x 36^4 (when number is at second). They all should be 10 x 36^5 why not? $\endgroup$ – Sanjay Apr 6 '17 at 9:08
  • $\begingroup$ The second case isn't "digit in second", it's "the first digit is in second", so you need a letter, then a digit, then anything. If you count "digit in first" then "digit in second", you've counted all the strings that start with two digits twice. $\endgroup$ – Especially Lime Apr 6 '17 at 9:13
  • $\begingroup$ @sanjay - the string 12AAAA would be counted by both of your cases, that is to say, double counted (or multiple counts) - try it with 3 character string, 1 letter and 1 number - you'll see how it double coubnts $\endgroup$ – Cato Apr 6 '17 at 9:13
  • $\begingroup$ @Cato That's the point I was missing. Thanks. $\endgroup$ – Sanjay Apr 6 '17 at 9:27

The number of upper case alphanumeric strings of length $6$ with at least one numeral is $$\sum_{k=1}^6\binom6k10^k26^{6-k}$$ where the term $\binom6k10^k26^{6-k}$ is the number of strings containing exactly $k$ numerals and $6-k$ upper case letters. Without the requirement of at least one numeral, the total number of strings is $$\sum_{k=0}^6\binom6k10^k26^{6-k}=(10+26)^6=36^6$$ by the binomial theorem, so the number of strings with at least one numeral is $$\sum_{k=1}^6\binom6k10^k26^{6-k}=\sum_{k=0}^6\binom6k10^k26^{6-k}-\binom6010^026^6=36^6-26^6.$$

Now, if you want the number of strings with at least two numerals, that is given "directly" by the sum $$\sum_{k=2}^6\binom6k10^k26^{6-k},$$ Or to make the calculation a little easier we can rewrite it as $$\sum_{k=0}^6\binom6k10^k26^{6-k}-\sum_{k=0}^1\binom6k10^k26^{6-k}=36^6-26^6-6\cdot10\cdot26^5.$$


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