Question

Find the number of odd integers between \[1000\] and \[8000\] which have none of their digit repeated.

Answer 1

Hint: We are provided two even integers \[1000\] and \[8000\] and we have to find odd integers between\[1000\] and \[8000\] which have none of their digits repeated.
An odd integer is those integers that end with \[1\], $3$ , $5$ , $7$ , $9$
Here we will use the fact that odd integer lying between \[1000\] and \[8000\] must be a \[3\]-digit number and unit digit will always end with \[1\], $3$ , $5$ , $7$ , $9$.

Complete step by step solution:
We are given a range \[1000\] \[ - \]\[8000\] both are even integers.
So, any odd integer lying between \[1000\] and \[8000\] neither has \[0\], $8$ , $9$ at its thousand places.
Then a thousand places can be filled in \[7\] ways.
Namely, any digit \[1\] to \[7\] the remaining two places can be filled in \[8 \times 7 = 56\] ways.
Hence Total no. formed in this way \[56 \times 7 = 392\]
Now if the unit place is filled with any of the four digits \[1\], $3$ , $5$ , $7$
 the thousand’s place can be filled in \[6\] ways and the remaining two places can be filled in \[8 \times 7 = 56\] ways
Now the total number formed in this way \[56 \times 6 \times 4 = 1344\]
The total number of odd integers formed will be \[392 + 1344 = 1736\] ways

Note: Students must avoid calculation mistakes to make their solutions error-free. Mistakes in calculation often lead them to incorrect and confusing answers.
In this question, the student must focus on the last place because they are supposed to find odd numbers. Therefore, the last digit matters here, then they need to focus on the first two places one by one.
Here in this question, one can also apply the permutation formula because permutation is also called arrangements and here we can see arrangement is formed because repetition is not allowed in the given question.