By using the non-zero digits, the number of 5 digit numbers that can be formed so that each number has largest digit in its middle place and the digits in the number are different is

Middle digit$=5\Rightarrow$$\mathrm{Remaining} 4 \mathrm{places} \mathrm{can} \mathrm{be} \mathrm{filled} \mathrm{using} 1, 2, 3, 4 \mathrm{is} 4{\mathrm{p}}_4 \mathrm{wa}ys$

$\mathrm{middle} \mathrm{digit}=6\Rightarrow \mathrm{Remaining} 4 \mathrm{places} \mathrm{can} \mathrm{be} \mathrm{filled} \mathrm{in} \mathrm{using} 1, 2, 3, 4, 5 \mathrm{in} 5{\mathrm{p}}_4 \mathrm{ways}.$

$\mathrm{Middle} \mathrm{digit}=7\Rightarrow \mathrm{using} 1, 2, 3, 4, 5, 6 \mathrm{remaining} 4 \mathrm{places} \mathrm{can} \mathrm{be} \mathrm{filled} \mathrm{in} 6{\mathrm{p}}_4 \mathrm{ways}.$

$\mathrm{Middle} \mathrm{dight}=8\Rightarrow \mathrm{using} 1, 2, 3, 4, 5, 6, 7 \mathrm{remaining} 4 \mathrm{places} \mathrm{cab} \mathrm{be} \mathrm{filled} \mathrm{in} 7{\mathrm{p}}_4 \mathrm{ways}.$

$\mathrm{Middle} \mathrm{digit}=9\Rightarrow \mathrm{using} 1, 2, 3, 4, 5, 6, 7, 8 \mathrm{remaining} 4 \mathrm{places} \mathrm{can} \mathrm{be} \mathrm{filled} \mathrm{in} 8{\mathrm{p}}_4 \mathrm{ways}.$

$\begin{array}{l}\therefore \mathrm{Total} \mathrm{no}. \mathrm{of} 5\mathrm{digit} \mathrm{numbers}=\sum _{r=4}^8\left({ }^r{P}_4\right)\\ \end{array}$