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$\left( a \right)\sum\limits_{n = 3}^9 {{}^n{P_4}} $

$\left( B \right)33\left( {3!} \right)$

$\left( c \right)30\left( {3!} \right)$

$\left( d \right)\sum\limits_{n = 3}^8 {n.{}^n{P_4}} $

Answer
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As we all know that there are a total 10 digits available which are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.

We have to make five digit numbers such that the middle digit is the largest.

So the possible middle digits are 4, 5, 6, 7, 8, and 9.

Case – 1: So when the middle digit (i.e. ${3^{rd}}$ place) is 4, remaining digits in a five digit number can be 0, 1, 2, 3.

Now we have to arrange them, as 0 digit cannot be at the first place so the number of ways to fill first place are 3 and the remaining ${2^{nd}},{4^{th}},{\text{ and }}{5^{th}}$ place can be filled by ${}^3{P_3}$.

So the number of ways such that the middle digit is 4 which is largest = $3 \times {}^3{P_3}$

Case – 2: now when the middle digit (i.e. ${3^{rd}}$ place) is 5, remaining digits in a five digit number can be 0, 1, 2, 3, 4.

Now we have to arrange them, as 0 digit cannot be at the first place so the number of ways to fill first place are 4 and the remaining ${2^{nd}},{4^{th}},{\text{ and }}{5^{th}}$ place can be filled by ${}^4{P_3}$.

So the number of ways such that the middle digit is 5 which is largest = $4 \times {}^4{P_3}$

Similarly for the remaining cases

Case – 3: When the middle digit is 6

So the number of ways such that the middle digit is 6 which is largest = $5 \times {}^5{P_3}$

Case – 4: When the middle digit is 7

So the number of ways such that the middle digit is 7 which is largest = $6 \times {}^6{P_3}$

Case – 5: When the middle digit is 8

So the number of ways such that the middle digit is 8 which is largest = $7 \times {}^7{P_3}$

Case – 6: When the middle digit is 9

So the number of ways such that the middle digit is 9 which is largest = $8 \times {}^8{P_3}$

So the total number of 5 digit numbers of different digits such that the middle digit is the largest is the sum of all the above cases.

$ \Rightarrow T = 3 \times {}^3{P_3} + 4 \times {}^4{P_3} + 5 \times {}^5{P_3} + 6 \times {}^6{P_3} + 7 \times {}^7{P_3} + 8 \times {}^8{P_3}$

We can also write this in a summation form so we have,

$ \Rightarrow T = \sum\limits_{n = 3}^8 {n \times {}^n{P_3}} $

So this is the required answer.

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