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Q.

What is the number of natural numbers from 1000 to 9999 (both inclusive) that do not have all four different digits?


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a

4464

b

5862

c

5825

d

4565 

answer is A.

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Detailed Solution

Concept- Let us first know about four – digit numbers.
FOUR – DIGIT NUMBERS:The figures that have four, i.e. 4 integers are hollered four – number figures. In distinct expressions, we can bring out that the figures that possess bones , knockouts, hundreds and
thousands of locations are called four –digit numbers.
The formula that's operated then's as follows
rnC=n!r!n-r! We require to detect the number of natural figures from 1000 to 9999( both inclusive) that serven't possess all four distinctive integers
Amount of numbers from 1000 to 9999
= (9999 – 1000) + 1  = 8999 + 1 
= 9000 Numbers that have all four different digits
=19C×19C×17C×17C
=9×9×8×7 = 4536 Amount of numbers that refrain from having all 4 different digits = 9000 – 4536
                                                                                             = 4464
Thus, the amount of numbers that refrain from having all 4 digits different is 4464.
Hence, the correct answer is option 1) 4464
 
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