30. Calculate: a) $C_{8}^{3}$; b) $C_{10}^{8}$.
Solution. a) Applying formula (6) for $m=8, n=3$, we find

$$
C_{8}^{3}=-\frac{8!}{(8-3)!3!}=-\frac{8!}{5!3!}=\frac{5!6 \cdot 7 \cdot 8}{5!3 \cdot 2 \cdot 1}=56
$$

b) $C_{10}^{\beta}=\frac{10!}{(10-8)!\cdot 8!}=\frac{10!}{2!\cdot 8!}=\frac{8!9 \cdot 10}{1 \cdot 2 \cdot 8!}=45$.

Let's note the main property of the number of combinations:

$$
C_{m}^{n}=C_{m}^{m-n}
$$

Indeed,

$$
C_{m}^{n}=\frac{m!}{(m-n)!n!}, C_{m}^{m-n}=\frac{m!}{(m-m+n)!(m-n)!}=\frac{m!}{n!(m-n)!}
$$

We see that the right-hand sides of these equalities are equal; therefore, the left-hand sides are also equal.

This property of the number of combinations allows simplifying the calculation of the number of combinations of $m$ elements taken $n$ at a time, when $n$ exceeds $\frac{1}{2} m$.