ข้อสอบ PAT 1 - มีนาคม 2555

$$\begin{gathered} \mathrm{จากโจทย์} \left\{\mathrm{a},\mathrm{b},\mathrm{c}\right\}\subset \mathrm{A} \mathrm{โดยที่} \mathrm{a}<\mathrm{b}<\mathrm{c} \\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{และ} \mathrm{a},\mathrm{b},\mathrm{c} \mathrm{เป็นลำดับเลขคณิต} \\ \mathbf{กรณี}\mathbf{ }\mathbf{1} \mathrm{d}=1 \\ \mathrm{จะได้} \left\{1,2,3\right\} , \left\{2,3,4\right\} , \left\{3,4,5\right\} , \left\{4,5,6\right\} , \left\{5,6,7\right\} \\ \mathbf{กรณี}\mathbf{ }\mathbf{2} \mathrm{d}=2 \\ \mathrm{จะได้} \left\{1,3,5\right\} , \left\{2,4,6\right\} , \left\{3,5,7\right\} \\ \mathbf{กรณี}\mathbf{ }\mathbf{3} \mathrm{d}=3 \\ \mathrm{จะได้} \left\{1,4,7\right\} \end{gathered}$$

$$\begin{gathered} \mathrm{แสดงว่า} \mathrm{n}\left(\mathrm{E}\right)=5+3+1 \\ =9 \mathrm{แบบ} \\ \mathrm{n}\left(\mathrm{S}\right)=\left(\begin{array}{c}7\\ 3\end{array}\right)=\frac{7!}{4!3!} \\ =35 \mathrm{แบบ} \end{gathered}$$

$$\begin{gathered} \mathrm{ดังนั้น} \mathrm{P}\left(\mathrm{E}\right)=\frac{\mathrm{n}\left(\mathrm{E}\right)}{\mathrm{n}\left(\mathrm{S}\right)} \\ =\frac{9}{35} \end{gathered}$$