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

Hint: To solve this question, we will find the number of natural numbers between 1000 and 9999 inclusive of these both. Then we will find the number of natural numbers formed where all the 4 digits are different. Then, to find the number of natural numbers from 1000 to 9999 that do not have all 4 different digits, we will subtract the number of 4-digit natural numbers with all unique digits from the number of all the natural numbers from 1000 to 9999.

Complete step-by-step answer:
Number of natural numbers between 1000 and 9999 are (9999 – 1000 + 1) = 9000.
Now, we will find all the 4 digits natural numbers with all unique digits.
The first place of the number cannot have 0, therefore, the number of possibilities of the first place is 9.
Now, if 1 of the 10 digits take the first place, 9 digits will be left. Thus, second place of the number can be occupied in 9 ways.
After 2 digits are decided, there will be 8 digits left to occupy the third place. They can be assigned in 8 ways.
This will leave us with 7 digits and the fourth place can be occupied in 7 ways.
Thus, the number of natural numbers = 9 $ \times $ 9 $ \times $ 8 $ \times $ 7 = 4536.
Therefore, 4536 natural numbers out of 9000 numbers have all the 4 digits that are unique. Therefore, rest numbers must have digits that are repeated.
Thus, the number of 4-digit numbers that do not have all 4 different digits is 9000 – 4536 = 4464.
Hence, option (b) is correct.

Note: Students are advised to be careful about the first place of the number, since it cannot be occupied by 0. It is possible to find the numbers directly, but it is very tedious. Hence, indirect approach is the best approach.