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

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$\sum_{r = 4}^{9} {}^{r}P_{4}$

$\sum_{r = 4}^{8} {}^{r}P_{4} - \sum_{r = 4}^{8} {}^{r}P_{3}$

$\sum_{r = 4}^{8} {}^{r}P_{3}$

$\sum_{r = 4}^{8} {}^{r}P_{4}$

answer is D.

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Detailed Solution

Middle digit $= 5 \Rightarrow$ $Remaining 4 places can be filled using 1, 2, 3, 4 is 4 p_{4} wa y s$

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

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

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

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

$\begin{array}{l} ∴ Total no . of 5 digit numbers = \sum_{r = 4}^{8} (^{r} P_{4}) \end{array}$

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