Now let's examine the Binomial Theorem and Multinomial theorem a little. I wouldn't worry too much about 6.25, it's more important to be able to apply the theorem, and in applying it you will gain a better understanding of the general idea anyway.

Let's start with the binomial theorem. As an example, the binomial theorem (similar to Pascal's triangle) says $$\begin{aligned} (x+y)^6 &= \binom{6}{0}x^6y^0 + \binom{6}{1}x^5y^1 + \binom{6}{2}x^4y^2 + \cdots + \binom{6}{6}x^0y^6 \\ &=\frac{6!}{6!0!}x^6y^0 + \frac{6!}{5!1!}x^5y^1 + \frac{6!}{4!2!}x^4y^2 + \cdots + \frac{6!}{0!6!}x^0y^6 \\ &= x^6 + 6 x^5y + 15 x^4y^2 + 20 x^3y^3 + 15x^2y^4 + 6xy^5 + y^6 \\ \end{aligned}$$ Examining the coefficients a little, note that the coefficient of $x^4y^2$ is $\dfrac{6!}{4!2!}$ which is the number of ways to arrange four $x$'s and two $y$'s. This is true for the other coefficients as well (and will continue to be true for the multinomial theorem).

To answer your question a bit from the other post, a lot of times when we say "use the binomial theorem" we really mean to plug in values for $x$ and $y$ in an equation like the one above. For example, if $x = 1$ and $y = 1$, note all powers of $x$ and $y$ disappear. Hence, we get $$2^6 = (1+1)^6 = \binom{6}{0} + \binom{6}{1} + \binom{6}{2} \cdots + \binom{6}{6}.$$ Thus, we've prove than $2^6$ equals the sum of these binomial coefficients! A similar argument using $x=1$ and $y = -1$ gives $$0 = (1+ (-1))^6 = \binom{6}{0} - \binom{6}{1} + \binom{6}{2} - \binom{6}{3} + \binom{6}{4} - \binom{6}{5} + \binom{6}{6}.$$ But wait, by symmetry, $\binom{6}{0} = \binom{6}{6}$, $\binom{6}{1} = \binom{6}{5}$, and $\binom{6}{2} = \binom{6}{4}$ so we actually have $$0 = 2\binom{6}{0} - 2 \binom{6}{1} + 2\binom{6}{2} - \binom{6}{3}$$ so actually $$\frac{1}{2}\binom{6}{3} = \binom{6}{0} - \binom{6}{1} + \binom{6}{2}.$$ Note this is exactly what we want for 6.30 when $n=1$. Try to do something similar for $n=2$, so you need to look at $(1+(-1))^{10}$.

I'll expand on this with the multinomial theorem a bit below when I have a chance.