By using the non-zero digits,the number of 5 | vedclass

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(D) Let the $5$ digit number be $d_1 d_2 d_3 d_4 d_5$. The middle digit is $d_3$. Since the digits must be distinct and non-zero,and $d_3$ is the larg

Vedclass · Accepted Answer

$\sum_{r=4}^8 { }^r P_4$

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(D) Let the $5$ digit number be $d_1 d_2 d_3 d_4 d_5$. The middle digit is $d_3$. Since the digits must be distinct and non-zero,and $d_3$ is the largest,$d_3$ must be at least $5$ (because we need $4$ distinct digits smaller than $d_3$). If $d_3 = 5$,the remaining $4$ digits must be chosen from ${1, 2, 3, 4}$. The number of ways to arrange these is ${ }^4 P_4$. If $d_3 = 6$,the remaining $4$ digits must be chosen from ${1, 2, 3, 4, 5}$. The number of ways to arrange these is ${ }^5 P_4$. If $d_3 = 7$,the remaining $4$ digits must be chosen from ${1, 2, 3, 4, 5, 6}$. The number of ways to arrange these is ${ }^6 P_4$. If $d_3 = 8$,the remaining $4$ digits must be chosen from ${1, 2, 3, 4, 5, 6, 7}$. The number of ways to arrange these is ${ }^7 P_4$. If $d_3 = 9$,the remaining $4$ digits must be chosen from ${1, 2, 3, 4, 5, 6, 7, 8}$. The number of ways to arrange these is ${ }^8 P_4$. Therefore,the total number of such numbers is ${ }^4 P_4 + { }^5 P_4 + { }^6 P_4 + { }^7 P_4 + { }^8 P_4 = \sum_{r=4}^8 { }^r P_4$.

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

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