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.
On
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 $$\sum_{k=0}^6\binom6k10^k26^{6-k}-\sum_{k=0}^1\binom6k10^k26^{6-k}=36^6-26^6-6\cdot10\cdot26^5.$$
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
$$10\times 36^5+26\times10\times36^4+26^2\times10\times36^3+26^3\times10\times36^2+26^4\times10\times36+26^5\times 10.$$