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This is textbook problem. Trying to prove by induction first for $n=1$ I get:

$$\frac{0!}{8!(0-8)!}+\frac{1!}{8!(1-8)!}=\frac{2!}{9!(2-9)!}$$

In all three fractions I have negative factorials, and factorials are as far as I know defined for positive integers only. What would $(-8)!$ be? $(-8)(-7)\cdots(-1)$?

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The formula $$\binom{n}{k} = \frac{n!}{k!(n-k)!}$$ only works if $n\geq k\geq 0$.

If $n < k$, then $\binom{n}{k} = 0.$ You can take this as a definition, or this can be deduced from the definition $$\binom{n}{k} = (\# \text{ of $k$-element subsets of an $n$-element set}).$$ If $n<k$, it is not possible for an $n$-element set of to have a $k$-element subset.

As a hint to your textbook problem, have you seen how the binomial coefficients $\binom{n}{k}$ fit together in Pascal's triangle?

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