We are given the values of the binomial coefficients:

$\binom{15}{8} = 6435, \quad \binom{16}{9} = 11440, \quad \binom{16}{10} = 8008$

We are tasked with finding $\binom{15}{10}$. Using Pascal's identity, we know:

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

Applying this to $\binom{16}{10}$, we have:

$\binom{16}{10} = \binom{15}{9} + \binom{15}{10}$

Thus,

$8008 = \binom{15}{9} + \binom{15}{10}$

Additionally, using Pascal's identity again on $\binom{16}{9}$:

$\binom{16}{9} = \binom{15}{8} + \binom{15}{9}$

Substituting the known values:

$11440 = 6435 + \binom{15}{9}$

Solving for $\binom{15}{9}$:

$\binom{15}{9} = 11440 - 6435 = 5005$

Now, substitute $\binom{15}{9} = 5005$ into the equation for $\binom{16}{10}$:

$8008 = 5005 + \binom{15}{10}$

Solving for $\binom{15}{10}$:

$\binom{15}{10} = 8008 - 5005 = 3003$

Thus, $\binom{15}{10} = 3003$.

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Would you like more details or have any questions? Here are some related questions you might find helpful:

1. How can Pascal’s identity be used to derive other binomial coefficients?
2. What is the significance of symmetry in binomial coefficients?
3. How can we express $\binom{n}{k}$ in terms of factorials?
4. How does Pascal’s triangle visually represent binomial coefficients?
5. Can we derive $\binom{16}{10}$ directly using factorials?

Tip: Symmetry in binomial coefficients means that $\binom{n}{k} = \binom{n}{n-k}$. This can simplify many computations!