$7 \cdot 71$ in the $(n, r)$ pairs where $0 \leqslant r \leqslant n \leqslant 63$
$$
C_{n}^{r}=\frac{n!}{r!(n-r)!}
$$

is even in how many pairs?
(Japan Mathematical Olympiad, 1991)
[Solution] As shown in the figure, take the modulus of each number in Pascal's triangle with $\bmod 2$, i.e., represent odd numbers as 1 and even numbers as 0.

Let the number of odd groups from the $2^{k}$-th row to the $2^{k+1}-1$-th row be denoted as $a_{k}$.
It is easy to see that, $\quad a_{k}=a_{k-1}+2 a_{k-1}=3 a_{k-1}$
Since $a_{1}=1, 63=2^{6}-1$ and $a_{k}=3^{k}$.
Therefore, the number of odd combinations in Pascal's triangle from the 0-th row to the 63-rd row is $a_{6}=$ $3^{6}=729$. Thus, the number of even combinations is $(1+2+\cdots+64)-729=1351$.