The question is as follows:
Four selections of single digits are made at random with replacement from the $10$ digits $0$ through $9$.
Find the probability that the sum of the selected digits will equal $20$.
Having a lot of difficulty with this problem, any help would be greatly appreciated.
Consider the polynomial $$p(x):=(1+x+x^2+\ldots+x^9)^4\ .$$ If we expand $p$ distributively then each selection of four digits in $[0..9]$ produces a term with coefficient $1$ in this expansion, and the exponent of that term is equal to the sum of the selected digits. It follows that the number $N$ of "good" selections is the coefficient of $x^{20}$ in this expansion, after terms have been collected.
We may rewrite $p$ as $$p(x)=\left({1-x^{10}\over 1-x}\right)^4=(1-4x^{10}+6x^{20}-\ldots)\sum_{k=0}^\infty{-4\choose k}(-x)^k\ .$$ Collecting terms we then obtain $$N={23\choose 20}-4{13\choose 10}+6=633\ .$$ The probability $P$ in question is therefore given by $$P={633\over 10^4}=6.33\%\ .$$