What is the greatest integer not exceeding the number $\left( 1 + \frac{\sqrt 2 + \sqrt 3 + \sqrt 4}{\sqrt 2 + \sqrt 3 + \sqrt 6 + \sqrt 8 + 4}\right)^{10}$?

$ 
\textbf{(A)}\ 2
\qquad\textbf{(B)}\ 10
\qquad\textbf{(C)}\ 21
\qquad\textbf{(D)}\ 32
\qquad\textbf{(E)}\ 36
$
1. We start with the given expression:
   \[
   \left( 1 + \frac{\sqrt{2} + \sqrt{3} + \sqrt{4}}{\sqrt{2} + \sqrt{3} + \sqrt{6} + \sqrt{8} + 4} \right)^{10}
   \]

2. First, simplify the denominator:
   \[
   \sqrt{2} + \sqrt{3} + \sqrt{6} + \sqrt{8} + 4
   \]

3. Notice that:
   \[
   \sqrt{8} = 2\sqrt{2} \quad \text{and} \quad \sqrt{4} = 2
   \]

4. Rewrite the denominator:
   \[
   \sqrt{2} + \sqrt{3} + \sqrt{6} + 2\sqrt{2} + 4
   \]

5. Combine like terms:
   \[
   3\sqrt{2} + \sqrt{3} + \sqrt{6} + 4
   \]

6. Now, consider the numerator:
   \[
   \sqrt{2} + \sqrt{3} + 2
   \]

7. We need to simplify the fraction:
   \[
   \frac{\sqrt{2} + \sqrt{3} + 2}{3\sqrt{2} + \sqrt{3} + \sqrt{6} + 4}
   \]

8. Notice that:
   \[
   (\sqrt{2} + 1)(\sqrt{2} + \sqrt{3} + 2) = 2 + \sqrt{6} + \sqrt{8} + \sqrt{2} + \sqrt{3} + 2
   \]

9. Simplify the product:
   \[
   2 + \sqrt{6} + 2\sqrt{2} + \sqrt{2} + \sqrt{3} + 2 = 3\sqrt{2} + \sqrt{3} + \sqrt{6} + 4
   \]

10. Therefore, the denominator can be rewritten as:
    \[
    (\sqrt{2} + 1)(\sqrt{2} + \sqrt{3} + 2)
    \]

11. The given expression simplifies to:
    \[
    \left(1 + \frac{1}{\sqrt{2} + 1}\right)^{10}
    \]

12. Simplify the fraction inside the parentheses:
    \[
    \frac{1}{\sqrt{2} + 1} = \frac{\sqrt{2} - 1}{(\sqrt{2} + 1)(\sqrt{2} - 1)} = \frac{\sqrt{2} - 1}{1} = \sqrt{2} - 1
    \]

13. Therefore, the expression becomes:
    \[
    \left(1 + \sqrt{2} - 1\right)^{10} = (\sqrt{2})^{10} = (2^{1/2})^{10} = 2^5 = 32
    \]

14. The greatest integer not exceeding 32 is 32.

The final answer is $\boxed{32}$