Question

What is the value of:

$\dfrac{1}{3} \cdot \dfrac{2}{4} \cdot \dfrac{3}{5} \cdots \dfrac{18}{20} \cdot \dfrac{19}{21} \cdot \dfrac{20}{22}$

$\displaystyle \dfrac{1}{462}$

$\displaystyle \dfrac{1}{231}$

$\displaystyle \dfrac{1}{132}$

$\displaystyle \dfrac{2}{213}$

$\displaystyle \dfrac{1}{22}$

Most numerator and denominator factors cancel

After cancellation, only $1\cdot2$ remains on top

Since every integer from $3$ to $20$ occurs once as a denominator and once as a numerator, they cancel each other out.

After canceling every number out, we have only $1$ and $2$ left as numerators and $21$ and $22$ left as denominators.

The remaining fraction is $\dfrac{1 \cdot 2}{21 \cdot 22} .$ This simplifies to $\dfrac{2}{462} = \dfrac{1}{231}$

Thus, the correct answer is B.