How many numbers between 100 and 1000 can be formed with the digits 5, 6, 7, 8, 9, if the repetition of digits is not allowed?
60
Let's break down how to solve this problem involving forming numbers with specific digits when repetition isn't allowed. We need to find how many three-digit numbers can be created using the digits 5, 6, 7, 8, and 9, ensuring that each digit is used at most once in a number.
Understanding the Problem: Three-Digit Numbers and Repetition
The question asks for numbers between 100 and 1000. This range includes all positive three-digit integers. A three-digit number has three places: the hundreds place, the tens place, and the units place.
We are given a set of 5 distinct digits: {5, 6, 7, 8, 9}.
The key constraint is that the repetition of digits is not allowed. This means that once a digit is used in one place (e.g., the hundreds place), it cannot be used again in the tens or units place for the same number.
Applying the Fundamental Counting Principle
We can determine the total number of possibilities by considering the choices for each digit place sequentially. This is a fundamental concept in combinatorics known as the Fundamental Counting Principle.
Let's consider the number of options for each position in the three-digit number:
- Hundreds Place: For the first digit (hundreds place), we can choose any of the 5 available digits (5, 6, 7, 8, or 9). So, there are 5 choices for the hundreds place.
- Tens Place: After selecting a digit for the hundreds place, we are left with 4 remaining digits. Since repetition is not allowed, the digit used in the hundreds place cannot be used here. Thus, there are 4 choices for the tens place.
- Units Place: After selecting digits for both the hundreds and tens places, we have used up 2 digits from the original set of 5. This leaves us with 3 remaining digits. So, there are 3 choices for the units place.
Calculating the Total Number of Numbers
To find the total number of distinct three-digit numbers that can be formed, we multiply the number of choices for each position:
Total number of numbers = (Choices for Hundreds Place) × (Choices for Tens Place) × (Choices for Units Place)
Total number of numbers = \(5 \times 4 \times 3\)
Total number of numbers = \(20 \times 3\)
Total number of numbers = \(60\)
We can represent this using a table:
| Digit Place | Number of Choices | Explanation |
|---|---|---|
| Hundreds | 5 | Any of the digits {5, 6, 7, 8, 9} |
| Tens | 4 | Any remaining digit (repetition not allowed) |
| Units | 3 | Any remaining digit (repetition not allowed) |
Relating to Permutations
This problem is also a classic example of a permutation because the order of the digits matters (e.g., 567 is different from 765), and we are selecting and arranging a subset of digits from a larger set without replacement (no repetition). The number of permutations of \(n\) distinct items taken \(r\) at a time is denoted by \(_nP_r\) and calculated as \(\frac{n!}{(n-r)!}\).
In this case, we have \(n=5\) distinct digits (5, 6, 7, 8, 9) and we are selecting and arranging \(r=3\) of them to form a three-digit number.
So, the number of possible three-digit numbers is:
\[ _5P_3 = \frac{5!}{(5-3)!} = \frac{5!}{2!} \] \[ _5P_3 = \frac{5 \times 4 \times 3 \times 2 \times 1}{2 \times 1} = 5 \times 4 \times 3 = 60 \]Both methods yield the same result, confirming our calculation.
Therefore, there are 60 numbers between 100 and 1000 that can be formed using the digits 5, 6, 7, 8, 9 without repeating any digit.
Revision Table: Counting Numbers
| Concept | Explanation | Application Here |
|---|---|---|
| Fundamental Counting Principle | If an event can occur in \(m\) ways and another independent event can occur in \(n\) ways, then both events can occur in \(m \times n\) ways. | Used to multiply the number of choices for each digit place (hundreds, tens, units). |
| Permutation | An arrangement of items where the order matters. | Choosing and arranging 3 distinct digits out of 5. Formula \(_nP_r\). |
| Repetition Not Allowed | Once an item is used, it cannot be used again in the same selection/arrangement. | The number of available choices decreases by one at each step of selecting a digit. |
Additional Information: Related Counting Problems
Let's explore how the problem changes with different constraints:
- If repetition of digits was allowed:
Hundreds place: 5 choices (5, 6, 7, 8, 9)
Tens place: 5 choices (any of the original digits can be repeated)
Units place: 5 choices (any of the original digits can be repeated)
Total numbers = \(5 \times 5 \times 5 = 5^3 = 125\)
- If the numbers had to be even and repetition was not allowed:
The available digits are {5, 6, 7, 8, 9}. The even digits are {6, 8}.
Units place: 2 choices (must be 6 or 8).
Hundreds place: 4 remaining choices (any of the 5 minus the one used for the units place).
Tens place: 3 remaining choices.
Total numbers = \(4 \times 3 \times 2 = 24\)
Understanding these variations helps solidify the concepts of permutations, combinations, and the fundamental counting principle in different scenarios.
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
