What is the largest integer $n$ for which $\binom{8}{3} + \binom{8}{4} = \binom{9}{n}$?
By Pascal's Identity, we have $\binom{8}{3} + \binom{8}{4} = \binom{9}{4}$.  However, we also have $\binom{9}{4} = \binom{9}{9-4} = \binom{9}{5}$.  There are no other values of $n$ such that $\binom{9}{4} = \binom{9}{n}$, so the largest possible value of $n$ is $\boxed{5}$.
5
Level 3
Counting & Probability
Problem: What is the largest integer $n$ for which $\binom{8}{3} + \binom{8}{4} = \binom{9}{n}$?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To solve the equation $\binom{8}{3} + \binom{8}{4} = \binom{9}{n}$, we first need to calculate the left-hand side. 

We start with the binomial coefficients. The value of $\binom{8}{3}$ can be computed as follows:
\[
\binom{8}{3} = \frac{8!}{3!(8-3)!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = \frac{336}{6} = 56.
\]

Next, we calculate $\binom{8}{4}$:
\[
\binom{8}{4} = \frac{8!}{4!(8-4)!} = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = \frac{1680}{24} = 70.
\]

Now, we add these two results together:
\[
\binom{8}{3} + \binom{8}{4} = 56 + 70 = 126.
\]

Next, we need to find $n$ such that $\binom{9}{n} = 126$. We can compute the binomial coefficients for $n = 0$ to $9$ to find the correct value.

Calculating $\binom{9}{0}$:
\[
\binom{9}{0} = 1.
\]

Calculating $\binom{9}{1}$:
\[
\binom{9}{1} = 9.
\]

Calculating $\binom{9}{2}$:
\[
\binom{9}{2} = \frac{9 \times 8}{2 \times 1} = 36.
\]

Calculating $\binom{9}{3}$:
\[
\binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84.
\]

Calculating $\binom{9}{4}$:
\[
\binom{9}{4} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126.
\]

At this point, we find that $\binom{9}{4} = 126$. 

Next, we also note that $\binom{9}{n}$ has the property that $\binom{9}{n} = \binom{9}{9-n}$. This means that $\binom{9}{5} = 126$ as well. 

Thus, the possible values for $n$ that satisfy the equation are $4$ and $5$. However, since we are looking for the largest integer $n$, we conclude that the largest integer $n$ for which $\binom{8}{3} + \binom{8}{4} = \binom{9}{n}$ is $5$.<|END_OF_LATENT|> The final answer is $5$.