An ellipse $$E : \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ passes through the vertices of the hyperbola $$H : \frac{x^2}{49} - \frac{y^2}{64} = -1$$. Let the major and minor axes of the ellipse $$E$$ coincide with the transverse and conjugate axes of the hyperbola $$H$$. Let the product of the eccentricities of $$E$$ and $$H$$ be $$\frac{1}{2}$$. If $$l$$ is the length of the latus rectum of the ellipse $$E$$, then the value of $$113l$$ is equal to______.

We need to compute $$\displaystyle\sum_{r=0}^{5} \dfrac{1}{\binom{11}{r}}$$ and express it as $$\dfrac{a}{b}$$ in lowest terms, then find $$a - b$$.

The binomial coefficients are: $$\binom{11}{0} = 1$$, $$\binom{11}{1} = 11$$, $$\binom{11}{2} = 55$$, $$\binom{11}{3} = 165$$, $$\binom{11}{4} = 330$$, $$\binom{11}{5} = 462$$.

The sum is $$\dfrac{1}{1} + \dfrac{1}{11} + \dfrac{1}{55} + \dfrac{1}{165} + \dfrac{1}{330} + \dfrac{1}{462}$$.

Finding the LCM of the denominators: $$\text{lcm}(1, 11, 55, 165, 330, 462)$$. We have $$462 = 2 \cdot 3 \cdot 7 \cdot 11$$, $$330 = 2 \cdot 3 \cdot 5 \cdot 11$$, $$165 = 3 \cdot 5 \cdot 11$$, $$55 = 5 \cdot 11$$. So $$\text{lcm} = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 = 2310$$.

Converting each fraction: $$\dfrac{2310}{1} + \dfrac{2310}{11} + \dfrac{2310}{55} + \dfrac{2310}{165} + \dfrac{2310}{330} + \dfrac{2310}{462}$$

$$= 2310 + 210 + 42 + 14 + 7 + 5 = 2588$$

So the sum is $$\dfrac{2588}{2310}$$. Simplifying: $$\gcd(2588, 2310)$$. $$2588 = 2310 + 278$$, $$2310 = 8 \cdot 278 + 86$$, $$278 = 3 \cdot 86 + 20$$, $$86 = 4 \cdot 20 + 6$$, $$20 = 3 \cdot 6 + 2$$, $$6 = 3 \cdot 2$$. So $$\gcd = 2$$.

$$\dfrac{2588}{2310} = \dfrac{1294}{1155}$$

Checking if this is in lowest terms: $$1294 = 2 \cdot 647$$, $$1155 = 3 \cdot 5 \cdot 7 \cdot 11$$. Since $$647$$ is prime and does not divide $$1155$$, $$\gcd(1294, 1155) = 1$$.

Therefore $$a = 1294$$, $$b = 1155$$, and $$a - b = 1294 - 1155 = 139$$.

The answer is $$139$$.