4. How many decimal numbers $\overline{a_{1} a_{2} a_{3} a_{4} a_{5} a_{6} a_{7} a_{8} a_{9} a_{10}}$, for which $a_{1}=1$ and each of the digits $a_{2}, a_{3}, \ldots, a_{10}$ is equal to 0 or 1, satisfy the condition

$$
a_{1}+a_{3}+a_{5}+a_{7}+a_{9}=a_{2}+a_{4}+a_{6}+a_{8}+a_{10} ?
$$
IV/4. If among the numbers $a_{3}, a_{5}, a_{7}$, and $a_{9}$ there are exactly $k$ numbers equal to 1, then among $a_{2}, a_{4}, a_{6}, a_{8}$, and $a_{10}$ there must be exactly $k+1$ numbers equal to 1. For some $k$, this can be done in $\binom{4}{k} \cdot\binom{5}{k+1}$ ways. Since $k \in\{0,1,2,3,4\}$, we have a total of $1 \cdot 5+4 \cdot 10+6 \cdot 10+4 \cdot 5+1 \cdot 1=126$ possibilities.

The realization that there must be exactly $k+1$ numbers equal to 1 among $a_{2}, a_{4}, a_{6}, a_{8}$, and $a_{10}$ if there are exactly $k$ numbers equal to 1 among $a_{3}, a_{5}, a_{7}$, and $a_{9}$, or equivalently: 3 points. The realization that there are $\binom{4}{k} \cdot\binom{5}{k+1}$ numbers in each case: 3 points. The calculation $1 \cdot 5+4 \cdot 10+6 \cdot 10+4 \cdot 5+1 \cdot 1=126: \mathbf{1}$ point.