Question

Find the number of eight-digit numbers the sum of whose digits is 4.

Answer 1

We need to find the number of 8-tuples ( $a_1,a_2,...,a_8$ ) of non-negative integers such that $a_1\ge 1$ and $a_1+a_2+...,+a_8$ = 4. If $a_1=1$.
There are three possibilities :
Either exactly three among $a_2,a_3,...,a_7$ equal 1 and the rest equal zero,
or
Five of them are zero and the other two equal 1 and 2,
or
Six of them are zero and the other equals 3.
In the rest case, there are $\left(\begin{array}{c}7\\ 3\end{array}\right)$ = 35 such 8-tuples,
In the second case there are $\left(\begin{array}{c}7\\ 2\end{array}\right)$ = 42 such 8-tuples
and in the third case there are 7 such 8-tuples.
If $a_1=2$ then either six of $a_2,a_3,..,a_7$ are zero and the other equals two, or five of them are zero and the remaining two both equal 1.
In the former case, there are 7 such 8-tuples and in the latter case there are $\left(\begin{array}{c}7\\ 2\end{array}\right)$ = 21 such 8-tuples.
If $a_1=3$ then exactly six of $a_2,a3,...,a_7$ are zero and the other equals one.
$\therefore$ 7 such 8-tuples.
Finally, there is one 8-tuple in which $a_1=4$ .
$\therefore$ there are 120 such 8-tuples.