Question
The total number of 5 digit numbers of different digits in which the digit in the middle is the largest, is
A.
$$\sum\limits_{n = 4}^9 {^n{P_4}} $$
B.
$$\sum\limits_{n = 4}^9 {^n{P_4}} - \frac{1}{{3!}}\sum\limits_{n = 3}^9 {^n{P_3}} $$
C.
$$30\left( {3!} \right)$$
D.
None of these
Answer :
None of these
Solution :
Since the largest digit is in the middle, the middle digit is greater than or equal to 4 the number of numbers with 4 in the middle $$ = {\,^4}{P_4} - {\,^3}{P_3}.$$
(∵ the other four places are to be filled by 0, 1, 2 and 3 and a number cannot begin with 0). Similarly, the numbers of numbers with 5 in the middle $$ = {\,^5}{P_4} - {\,^4}{P_3},{\text{ e}}{\text{.t}}{\text{.c}}{\text{.)}}$$
∴ The required number of numbers
$$\eqalign{
& = \left( {^4{P_4} - {\,^3}{P_3}} \right) + \left( {^5{P_4} - {\,^4}{P_3}} \right) + \left( {^6{P_4} - {\,^5}{P_3}} \right) + ..... + \left( {^9{P_4} - {\,^8}{P_3}} \right) \cr
& = \sum\limits_{n = 4}^9 {^n{P_4} - } \sum\limits_{n = 3}^8 {^n{P_3}} \cr} $$