In a certain chess tournament, after the 7th round (i.e., 7 games), a player has 5 points. In how many ways could this result have been achieved? (A win is 1 point, a draw is $1 / 2$ point, a loss is 0 points.)
Out of 7 games, 5 points can clearly only be achieved in the following ways: a) 5 wins, 2 losses, b) 4 wins, 2 draws, 1 loss, or c) 3 wins, 4 draws.

In case a), we are talking about the permutations of the elements $1,1,1,1,1,0,0$.

Their number

$$
P_{7}^{5,2}=\frac{7!}{5!2!}=21
$$

We can also reason as follows: any non-repeating second-class combination of the elements I, II, ..., VII can indicate the two rounds on which the losses occurred. Therefore, the number of possible cases is

$$
C_{7}^{2}=\binom{7}{2}=\frac{7 \cdot 6}{1 \cdot 2}=21
$$

In case b), with the same reasoning,

$$
P_{7}^{4,2}=\frac{7!}{4!2!}=105
$$

or

$$
C_{7}^{4} \cdot C_{3}^{2}=\binom{7}{4}\binom{3}{2}=\binom{7}{3}\binom{3}{1}=35 \cdot 3=105
$$

(If we switch the order of selection in the latter case, we naturally arrive at the same result: $C_{7}^{2} \cdot C_{5}^{4}=$ $\left.\binom{7}{2}\binom{5}{4}=21 \cdot 5=105.\right)$

In case c), similarly,

$$
P_{7}^{3,4}=\frac{7!}{4!3!}=35
$$

or

$$
C_{7}^{3}=\binom{7}{3}=35
$$

Therefore, the desired result can occur in $21+105+35=161$ ways.