Let $$\beta(\mathrm{m}, \mathrm{n})=\int_\limits0^1 x^{\mathrm{m}-1}(1-x)^{\mathrm{n}-1} \mathrm{~d} x, \mathrm{~m}, \mathrm{n}>0$$. If $$\int_\limits0^1\left(1-x^{10}\right)^{20} \mathrm{~d} x=\mathrm{a} \times \beta(\mathrm{b}, \mathrm{c})$$, then $100(\mathrm{a}+\mathrm{b}+\mathrm{c})$ equals _________.

<p>First, let's rewrite the given integral using the given form of the Beta function. The given integral is:</p> <p> <p>$\int_\limits0^1\left(1-x^{10}\right)^{20} \mathrm{~d} x$</p> </p> <p>To use the Beta function, let us make a substitution. Let $ x^{10} = t $. Then, $ dx = \frac{1}{10}t^{-\frac{9}{10}} dt $ or $ dx = \frac{1}{10} t^{-\frac{9}{10}} dt $. The limits of integration change as follows: when $ x = 0 $, $ t = 0 $, and when $ x = 1 $, $ t = 1 $.</p> <p>Substituting these into the integral, we have:</p> <p> <p>$\int_\limits0^1 (1 - t)^{20} \cdot \frac{1}{10} t^{-\frac{9}{10}} dt$</p> </p> <p>which simplifies to:</p> <p> <p>$\frac{1}{10} \int_\limits0^1 (1 - t)^{20} t^{-\frac{9}{10}} dt$</p> </p> <p>We recognize this integral as a Beta function $ \beta(m, n) $ where $ m = 1 - \frac{9}{10} = \frac{1}{10} $ and $ n = 20 + 1 = 21 $.</p> <p>Therefore, we can write this as:</p> <p> <p>$\frac{1}{10} \beta \left( \frac{1}{10}, 21 \right)$</p> </p> <p>Comparing this to $ a \times \beta(b, c) $, we have $ a = \frac{1}{10} $, $ b = \frac{1}{10} $, and $ c = 21 $.</p> <p>Now we calculate $ 100(a + b + c) $:</p> <p> <p>$$100 \left( \frac{1}{10} + \frac{1}{10} + 21 \right) = 100 \left( \frac{1}{10} + \frac{1}{10} + 21 \right) = 100 \left( \frac{1}{5} + 21 \right) = 100 \left( \frac{1}{5} + \frac{105}{5} \right) = 100 \left( \frac{106}{5} \right) = 100 \times 21.2 = 2120$$</p> </p> <p>So, the answer is Option D, 2120.</p>