The number of 3-digit even numbers that can be formed from the digits 0, 1, 2, 3, 4 and 5, repetition of digits being not allowed, is
52
Calculating 3-Digit Even Numbers Without Repetition
The problem asks us to find the total number of 3-digit even numbers that can be formed using the digits 0, 1, 2, 3, 4, and 5, with the condition that no digit is repeated in a number.
We are forming a 3-digit number, which has a hundreds place, a tens place, and a units place. For the number to be even, the digit in the units place must be an even digit from the given set. The even digits available are 0, 2, and 4.
Also, a 3-digit number cannot have 0 in the hundreds place.
Since repetition of digits is not allowed and the digit 0 has a special restriction (cannot be in the hundreds place), it is best to solve this problem by considering two cases based on the digit in the units place:
- Case 1: The units digit is 0.
- Case 2: The units digit is a non-zero even digit (2 or 4).
Case 1: The units digit is 0
In this case, the units place is fixed with the digit 0.
- Units Place: There is only 1 choice (0).
Now, we need to fill the hundreds and tens places using the remaining digits without repetition. The available digits are {0, 1, 2, 3, 4, 5}. Since 0 is used in the units place, the remaining digits are {1, 2, 3, 4, 5}.
- Hundreds Place: Cannot be 0 (already used). We have 5 remaining digits {1, 2, 3, 4, 5} to choose from. So, there are 5 choices for the hundreds place.
After filling the hundreds place, we have used two distinct digits (0 in units, and one from {1, 2, 3, 4, 5} in hundreds). There are 6 total digits. So, \(6 - 2 = 4\) digits remain for the tens place.
- Tens Place: Can be any of the remaining 4 digits. So, there are 4 choices for the tens place.
Total number of 3-digit even numbers when the units digit is 0 is calculated as:
\( \text{Number of choices for Hundreds} \times \text{Number of choices for Tens} \times \text{Number of choices for Units} \)
\( 5 \times 4 \times 1 = 20 \)
There are 20 such numbers.
Case 2: The units digit is a non-zero even digit (2 or 4)
In this case, the units place can be filled with either 2 or 4.
- Units Place: There are 2 choices (2 or 4).
Now, we need to fill the hundreds and tens places using the remaining digits without repetition. Let's say we picked one digit for the units place (e.g., 2). The available digits are {0, 1, 2, 3, 4, 5}. One non-zero digit (2 or 4) is used in the units place.
- Hundreds Place: Cannot be 0 (as it's a 3-digit number) and cannot be the digit used in the units place. We started with 6 digits. 0 is excluded for the hundreds place (1 restriction). The units digit is also excluded (1 restriction). So, we have \(6 - 1 - 1 = 4\) choices for the hundreds place.
After filling the hundreds and units places, we have used two distinct digits. There are 6 total digits. So, \(6 - 2 = 4\) digits remain for the tens place.
- Tens Place: Can be any of the remaining 4 digits. So, there are 4 choices for the tens place.
Total number of 3-digit even numbers when the units digit is 2 or 4 is calculated as:
\( \text{Number of choices for Hundreds} \times \text{Number of choices for Tens} \times \text{Number of choices for Units} \)
\( 4 \times 4 \times 2 = 32 \)
There are 32 such numbers.
Total Number of 3-Digit Even Numbers
The total number of 3-digit even numbers is the sum of the numbers from Case 1 and Case 2.
\( \text{Total} = \text{Numbers with Units digit 0} + \text{Numbers with Units digit 2 or 4} \)
\( \text{Total} = 20 + 32 = 52 \)
Thus, there are 52 three-digit even numbers that can be formed from the digits 0, 1, 2, 3, 4, and 5 without repetition.
Revision Table: 3-Digit Even Numbers
| Place Value | Case 1 (Units = 0) | Case 2 (Units = 2 or 4) |
|---|---|---|
| Units | 1 choice (0) | 2 choices (2 or 4) |
| Hundreds | 5 choices (from remaining {1,2,3,4,5}) | 4 choices (cannot be 0 or the chosen units digit) |
| Tens | 4 choices (from remaining 4 digits) | 4 choices (from remaining 4 digits) |
| Total for Case | \(5 \times 4 \times 1 = 20\) | \(4 \times 4 \times 2 = 32\) |
Total = \(20 + 32 = 52\).
Additional Information: Permutations and Combinations
This problem involves the concept of permutations because the order of the digits matters (e.g., 124 is different from 142). When repetition is not allowed, we are essentially counting permutations of distinct objects under certain constraints.
- Permutation: An arrangement of objects in a specific order. The number of permutations of \(n\) distinct objects taken \(r\) at a time is denoted by \(P(n, r)\) or \(^nP_r\), and is calculated as \(P(n, r) = \frac{n!}{(n-r)!}\).
- Combination: A selection of objects where the order does not matter. The number of combinations of \(n\) distinct objects taken \(r\) at a time is denoted by \(C(n, r)\) or \(^nC_r\), and is calculated as \(C(n, r) = \frac{n!}{r!(n-r)!}\).
In problems like forming numbers or arranging letters, the order is important, so permutations or the fundamental counting principle (as used above by considering choices for each position) are applicable.
Constraints like "even number" or "cannot start with 0" require careful consideration and often necessitate breaking the problem into separate cases, as demonstrated in the solution.
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 ?
A polygon has 44 diagonals then the number of its sides is
The number of ways in which 3-holiday tickets can be given to 20 employees of an organization if each employee is eligible for any one or more of the tickets, is
The number of ways in which a cricket team of 11 players can be chosen out of a batch of 15 players so that the captain of the team is always included, is
