112.解析：(1)在四列中分别取一格，分别取第一、二、三、四行中的某一格，即相当于把取出的格子排序.故A_{4}^{4}=24种选法.四、解答题：本题共5小题，共77分.解答应写出文字说明、证明过程或演算步骤.15.(13分)记\triangle ABC的内角A，B，C的对边分别为a，已知\sin A+\sqrt {3} \cos A=2.(1)求A.(2)若a=2, \sqrt {2}b \sin C=c \sin 2B,求\triangle ABC的周长.解：.\cdot \sin A+\sqrt {3} \cos A=2, \therefore 2 \sin(A+\frac { \pi }{3})=2, \sin(A+\frac { \pi }{3})=1.又A \in(0, \pi), \therefore A+\frac { \pi }{3}= \frac { \pi }{2},A= \frac { \pi }{6}.综上，角A为\frac { \pi }{6}.(2)\because \sqrt {2}b \sin c=c \sin 2B, \therefore \sqrt {2}b \sin C=2c \sin B \cos B, \therefore \sqrt {2}bc=2bc \cos B, \cos B= \frac { \sqrt {2}}{2}.XB \in(0, \pi), \therefore B= \frac { \pi }{4},C= \pi -A-B= \frac {7}{12} \pi.在\triangle ABC中，由正弦定理得\frac {a}{ \sin A}= \frac {b}{ \sin B}= \frac {c}{ \sin C}= \frac {2}{ \frac {1}{2}}=4,:b=4 \sin B=2 \sqrt {2},c=4 \sin C=4 \sin \frac {7 \pi }{12}=4 \sin(\frac { \pi }{4}+\frac { \pi }{3})= \sqrt {6}+\sqrt {2}.\therefore a+综上，\triangle ABC的周长为2+3 \sq