Counting Odd Integers with No Repeated Digits

The problem asks us to find the number of odd integers between 1000 and 9999 that have no repeated digits. Numbers between 1000 and 9999 are 4-digit numbers.

Let the 4-digit number be represented by the digits $abcd$, where $a$ is the thousands digit, $b$ is the hundreds digit, $c$ is the tens digit, and $d$ is the units digit.

We need to satisfy the following conditions:

• The number is a 4-digit number: This means the thousands digit $a$ cannot be 0. So, $a \in \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$.
• The number is odd: This means the units digit $d$ must be an odd digit. So, $d \in \{1, 3, 5, 7, 9\}$.
• No digit is repeated: All four digits $a, b, c, d$ must be distinct.

Step-by-Step Calculation for No Repeated Digits

To count these numbers, it's easiest to start with the positions that have the most restrictions. The units digit ($d$) has a restriction (must be odd), and the thousands digit ($a$) has a restriction (cannot be 0 and must be distinct from $d$).

Let's fill the positions step by step:

1. Units Digit ($d$): The number must be odd, so the units digit $d$ must be one of {1, 3, 5, 7, 9}. There are 5 choices for $d$.
2. Thousands Digit ($a$): The thousands digit $a$ cannot be 0. Also, the digits cannot be repeated, so $a$ cannot be the digit chosen for $d$.
   • Total digits available are {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} (10 digits).
   • One digit is used for $d$. Remaining 9 digits.
   • From these 9 digits, 0 is not allowed for $a$.
   • So, the number of choices for $a$ is the total remaining digits minus the digit 0. That is $9 - 1 = 8$ choices for $a$.
3. Hundreds Digit ($b$): The digit $b$ cannot be the digit chosen for $a$ and cannot be the digit chosen for $d$, as no digits can be repeated.
   • Two digits have been used (for $a$ and $d$).
   • Out of the 10 total digits, $10 - 2 = 8$ digits remain.
   • Any of these 8 remaining digits can be used for $b$. There are 8 choices for $b$.
4. Tens Digit ($c$): The digit $c$ cannot be the digit chosen for $a$, $d$, or $b$, as no digits can be repeated.
   • Three digits have been used (for $a$, $d$, and $b$).
   • Out of the 10 total digits, $10 - 3 = 7$ digits remain.
   • Any of these 7 remaining digits can be used for $c$. There are 7 choices for $c$.

To find the total number of such integers, we multiply the number of choices for each position:

$\backslash text\{Total\; number\; of\; integers\}\; =\; (\backslash text\{choices\; for\; \}\; d)\; \backslash times\; (\backslash text\{choices\; for\; \}\; a)\; \backslash times\; (\backslash text\{choices\; for\; \}\; b)\; \backslash times\; (\backslash text\{choices\; for\; \}\; c)$

$\backslash text\{Total\; number\; of\; integers\}\; =\; 5\; \backslash times\; 8\; \backslash times\; 8\; \backslash times\; 7$

$\backslash text\{Total\; number\; of\; integers\}\; =\; 40\; \backslash times\; 56$

$\backslash text\{Total\; number\; of\; integers\}\; =\; 2240$

Therefore, there are 2240 odd integers between 1000 and 9999 with no digit repeated.

Understanding Digit Constraints and Counting

This type of problem involves permutations with restrictions. We are essentially arranging distinct digits (no repetition) while satisfying conditions on the first and last digits (range and oddness). The constraints require careful consideration of which digits are available at each step.

Total = $5 \times 8 \times 8 \times 7 = 2240$.

Revision Table: Key Concepts

Additional Information: Permutations vs. Combinations

This problem involves permutations because the order of the digits matters (e.g., 1235 is different from 5321). A combination problem would be asking for a set of digits without regard to their arrangement.

When solving counting problems with restrictions, it is often helpful to address the restrictions first. In this case, the restrictions were on the units digit (odd) and the thousands digit (non-zero and distinct). By filling these positions first, we simplify the counting for the remaining positions.

The number of available digits decreases with each position filled because repetition is not allowed. This is characteristic of permutation problems where elements are chosen from a set without replacement.