Question

The number of natural numbers from $1000$ to $9999$ ( both inclusive) that do not have all $4$ different digits is

• A. $4048$
• B. $4464$
• C. $4518$
• D. $4536$

Answer 1

Answer: (B)

$4464$

There are total $9000$ numbers between $1000$ and $9999$

We will find the numbers which have all different digits and subtract them from $9000$

The number which have all numbers different=
$A\to 9$ $(choice1,2,3,4,5,6,7,8,9)$

$B\to 9$ (we can't select the number which is assigned to $A$)

$C\to 8$ (we can't select the number which is assigned to $A$ and $B$)

$D\to 7$ (we can't select the number which is assigned to $A$, $B$ and $C$)

Total$=9\times 9\times 8\times 7=4536$

required number of natural number$=9000-4536=4464.$