47. If $\mathrm{A}=2011^{2011}, B=(1 \times 2 \times \cdots \times 2011)^{2}$, then $\mathrm{A}$ $\qquad$ B. (Fill in “ $>$ "“=” or “＜")
Solution: $\frac{A}{B}=\frac{2011 \times 2011 \times \cdots \times 2011}{1 \times 2 \times \cdots \times 2011 \times 2011 \times \cdots \times 2 \times 1}=\frac{2011}{1 \times 2011} \times \frac{2011}{2 \times 2010} \times \cdots \times \frac{2011}{k \times(2011-k+1)} \times \cdots \times \frac{2011}{2011 \times 1}$,

Since when the sum of two numbers is constant, their product is smaller as their difference increases, the denominators in the above expression are no less than 2011, while the numerator is 2011, so, $\mathrm{A}<\mathrm{B}$.