Question

The total number of five-digit numbers of different digits in which the digit in the middle is the largest is
$\left( a \right)\sum\limits_{n = 3}^9 {{}^n{P_4}}$
$\left( B \right)33\left( {3!} \right)$
$\left( c \right)30\left( {3!} \right)$
$\left( d \right)\sum\limits_{n = 3}^8 {n.{}^n{P_4}}$

Answer 1

In this particular question use the concept that in a five digit number if the middle digit is largest so first possible middle digit is 4 because the remaining 4 places can be fill by only 0, 1, 2, and 3 and use the concept that 0 cannot be on the first place so the number of ways to arrange first place when middle digit is 4 are 3 and the remaining places can be filled by ${}^3{P_3}$, so use these concepts to reach the solution of the question.

Complete step by step answer:
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.

So, the correct answer is “Option D”.

Note: Whenever we face such types of questions the key concept we have to remember is that the number of ways to arrange n different objects in r places is given as ${}^n{P_r}$, so first find out all the cases as above then the total number of ways is the sum of all the cases so add them as above and simplify we will get the required answer.