How many four-digit natural numbers are there such that all of the digits are odd ?
625
Finding Four-Digit Natural Numbers with Only Odd Digits
We are asked to find the count of four-digit natural numbers where all the digits are odd numbers. A four-digit natural number is an integer between 1000 and 9999, inclusive.
Understanding the Problem Constraints
- The number must have exactly four digits.
- The number must be a natural number (a positive integer).
- Every digit in the number must be an odd digit.
Identifying the Available Digits
The odd digits are 1, 3, 5, 7, and 9. There are 5 odd digits available for use in constructing the four-digit numbers.
Applying the Counting Principle
A four-digit number has four positions for digits:
- Thousands digit
- Hundreds digit
- Tens digit
- Units digit
Since all digits must be odd, we need to determine how many choices we have for each position.
- Thousands Digit: The thousands digit must be an odd digit. The available odd digits are {1, 3, 5, 7, 9}. So, there are 5 choices for the thousands digit.
- Hundreds Digit: The hundreds digit must also be an odd digit. The available odd digits are {1, 3, 5, 7, 9}. So, there are 5 choices for the hundreds digit.
- Tens Digit: The tens digit must also be an odd digit. The available odd digits are {1, 3, 5, 7, 9}. So, there are 5 choices for the tens digit.
- Units Digit: The units digit must also be an odd digit. The available odd digits are {1, 3, 5, 7, 9}. So, there are 5 choices for the units digit.
Since the choice of digit for each position is independent of the choices for the other positions, we can use the multiplication principle to find the total number of such four-digit numbers.
Calculating the Total Number of Possibilities
The total number of four-digit natural numbers with all odd digits is the product of the number of choices for each digit position.
Total numbers = (Choices for Thousands Digit) × (Choices for Hundreds Digit) × (Choices for Tens Digit) × (Choices for Units Digit)
Total numbers = $5 \times 5 \times 5 \times 5$
Total numbers = $5^4$
Now, we calculate the value of $5^4$:
$5^4 = 5 \times 5 \times 5 \times 5 = 25 \times 25 = 625$
Therefore, there are 625 four-digit natural numbers such that all of the digits are odd.
Summary of Choices per Digit
| Digit Position | Available Digits | Number of Choices |
|---|---|---|
| Thousands | {1, 3, 5, 7, 9} | 5 |
| Hundreds | {1, 3, 5, 7, 9} | 5 |
| Tens | {1, 3, 5, 7, 9} | 5 |
| Units | {1, 3, 5, 7, 9} | 5 |
Total = $5 \times 5 \times 5 \times 5 = 625$.
The final answer is 625.
Important Questions from Permutations and Combinations
What is the number of 6-digit numbers that can be formed only by using 0, 1, 2, 3, 4 and 5 (each once); and divisible by 6 ?
Consider the following statements for a fixed natural number n:
1. C(n, r) is greatest if n = 2r
2. C(n, r) is greatest if n = 2r - 1 and n = 2r + 1
Which of the statements given above is/are correct ?
Let x be the number of permutations of the word ‘PERMUTATIONS’ and y be the number of permutations of the word ‘COMBINATIONS’. Which one of the following is correct ?
What is the number of ways in which 3 holiday travel tickets are to be given to 10 employees of an organization, if each employee is eligible for any one or more of the tickets?
A polygon has 44 diagonals then the number of its sides is
