Given that $\binom{23}{3}=1771$, $\binom{23}{4}=8855$, and $\binom{23}{5}=33649$, find $\binom{25}{5}$.
We can use Pascal's identity $\binom{n-1}{k-1}+\binom{n-1}{k}=\binom{n}{k}$ to find $\binom{24}{4}$ and $\binom{24}{5}$.

$$\binom{24}{4}=\binom{23}{3}+\binom{23}{4}=1771+8855=10626$$ $$\binom{24}{5}=\binom{23}{4}+\binom{23}{5}=8855+33649=42504$$

Now that we have $\binom{24}{4}$ and $\binom{24}{5}$, we can use Pascal's identity again to find $\binom{25}{5}$.

$$\binom{25}{5}=\binom{24}{4}+\binom{24}{5}=10626+42504=\boxed{53130}$$
53130
Level 2
Counting & Probability
Problem: Given that $\binom{23}{3}=1771$, $\binom{23}{4}=8855$, and $\binom{23}{5}=33649$, find $\binom{25}{5}$.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find $\binom{25}{5}$, we can use the identity that relates binomial coefficients: 

$$\binom{n}{k} = \binom{n-1}{k} + \binom{n-1}{k-1}.$$

We will apply this identity to express $\binom{25}{5}$ in terms of binomial coefficients with $n=24$ and $n=23$.

First, we calculate $\binom{24}{5}$:

$$\binom{24}{5} = \binom{23}{5} + \binom{23}{4}.$$

We already know $\binom{23}{5} = 33649$ and $\binom{23}{4} = 8855$. Thus,

$$\binom{24}{5} = 33649 + 8855 = 42504.$$

Next, we calculate $\binom{24}{4}$:

$$\binom{24}{4} = \binom{23}{4} + \binom{23}{3}.$$

Using the known values, we have:

$$\binom{24}{4} = 8855 + 1771 = 10626.$$

Now we can find $\binom{25}{5}$:

$$\binom{25}{5} = \binom{24}{5} + \binom{24}{4}.$$

Substituting the values we calculated:

$$\binom{25}{5} = 42504 + 10626 = 53130.$$

Thus, the final answer is $53130$.<|END_OF_LATENT|> The final answer is $53130$.