Problem source

$47231$ is a five-digit number whose digits sum to $4+7+2+3+1 = 17\,$.
\begin{questionparts}
\item Show that there are $15$ five-digit numbers whose digits sum to $43$. 
You should explain  your reasoning clearly.
\item How many five-digit numbers are there whose digits sum to $39$?
\end{questionparts}

Solution source

\begin{questionparts}
\item The largest a five-digit number can have for its digit sum is $45 = 9+9+9+9+9$. To achieve $43$ we can either have 4 9s and a 7 or 3 9s and 2 8s. The former can be achieved in $5$ ways and the latter can be achieved in $\binom{5}{2} = 10$ ways. (2 places to choose to put the 2 8s). In total this is $15$ ways.

\item To achieve $39$ we can have:

\begin{array}{c|l|c}
\text{numbers} & \text{logic} & \text{count} \\ \hline 
99993 & \binom{5}{1} & 5 \\
99984 & 5 \cdot 4 & 20 \\
99974 & 5 \cdot 4 & 20 \\
99965 & 5 \cdot 4 & 20 \\
99884 & \binom{5}{2} \binom{3}{2} & 30 \\
99875 & \binom{5}{2} 3! & 60 \\
99866 & \binom{5}{2} \binom{3}{2} & 30 \\
98886 & 5 \cdot 4 & 20 \\
98877 & \binom{5}{2} \binom{3}{2} & 30 \\
88887 & \binom{5}{1} & 5 \\ \hline 
&& 240
\end{array}

\end{questionparts}