None of the other problems posted so far that I have read have answered my question.
I have been able to solve the specific problems but I am seeking the generalisation for the two following problems
Setup: we have a license plate with a sequence of 3 digits ranging from 0-9 ( 10 possibilies each spot ).
1) find p(sum of digits = 25). I wrote the 6 possible permutations to get 6/1000. How do I generalise this to say sequence of 5 digits = 33? I have not made much success on generalising this.
2) p( the first digit is bigger than the third). Here I got 45/100. I believe I found a generalisation by making a grid of lattice points n*n. And I get n(n-1)/2.
Thanks for taking the time to read my post
Your first problem could be generalized using generating functions: We find the coefficient of the $z^{25}$ term in
$$ (1+z+z^2+\cdots+z^8+z^9)^3 $$
This essentially "counts up" the sum of the digits in the degree of each $z$-term, and the coefficient counts up the number of ways that sum can be obtained.
Your second problem can be conceived of more generally by symmetry. There are three possibilities: the first digit is less than the third digit, the first digit is more than the third digit, or the first digit is equal to the third digit. Since there is no preference for either digit, the probability that the first digit is more than the third digit should be one-half the probability that they aren't equal. Since that latter probability is $1/10$, the probability that the first digit is more than the third digit should be
$$ \frac12 \left( 1-\frac{1}{10} \right) = \frac{9}{20} $$