How many 3 digit odd numbers can be made from the digits 5, 2, 1, 3, 4 ? (without repetiton)
36
Number Formation Basics
The problem asks us to find the number of 3-digit odd numbers that can be formed using the digits 5, 2, 1, 3, and 4 without repeating any digit. A 3-digit number has three places: the hundreds place, the tens place, and the units place.
For a number to be odd, its units digit must be an odd number. The given digits are 5, 2, 1, 3, 4. Let's identify the odd digits from this set.
- Odd digits in the set {5, 2, 1, 3, 4} are 1, 3, and 5.
- Even digits in the set {5, 2, 1, 3, 4} are 2 and 4.
There are a total of 5 available digits.
Counting the Possibilities
Since the number must be a 3-digit odd number and digits cannot be repeated, we should fill the places considering the constraints. It is usually easiest to start with the most constrained position, which is the units place in this case (it must be odd).
Units Place
The units place must be filled by an odd digit. The available odd digits are 1, 3, and 5. So, there are 3 choices for the units place.
Number of choices for Units Place = 3 (1, 3, or 5)
Hundreds Place
Now, consider the hundreds place. We have 5 original digits. One digit has been used to fill the units place. Since repetition is not allowed, we have 5 - 1 = 4 digits remaining to choose from for the hundreds place.
Number of choices for Hundreds Place = 4
Tens Place
Finally, consider the tens place. We started with 5 digits. One digit was used for the units place, and another was used for the hundreds place. So, we have 5 - 2 = 3 digits remaining to choose from for the tens place.
Number of choices for Tens Place = 3
Total Number of 3-Digit Odd Numbers
To find the total number of 3-digit odd numbers that can be formed, we multiply the number of choices for each place:
\(\text{Total numbers} = (\text{Choices for Hundreds}) \times (\text{Choices for Tens}) \times (\text{Choices for Units})\)
\(\text{Total numbers} = 4 \times 3 \times 3\)
\(\text{Total numbers} = 12 \times 3\)
\(\text{Total numbers} = 36\)
So, there are 36 possible 3-digit odd numbers that can be formed using the digits 5, 2, 1, 3, 4 without repetition.
Let's summarize the choices in a table:
| Place | Constraint | Number of Choices |
|---|---|---|
| Units | Must be odd (1, 3, or 5) | 3 |
| Hundreds | Any remaining digit | 4 |
| Tens | Any of the last 3 remaining digits | 3 |
Total number of odd numbers = \(3 \times 4 \times 3 = 36\).
Important Questions from Permutation and Combination
The following figure is made with nine squares of length 1 cm each.
A student is standing at point P. The total number of shortest paths to reach point Q from point P, if the student walks along the boundaries of the squares only, is:
Consider the following statements (all the angles are measured in degrees) :
1. If two supplementary angles are $4x + 40$ and $10x$, then $x$ is;
2. The complementary angle of $\frac{4}{5} \times 45$ is :
3. If two complementary angles are in the ratio of $4 : 5$, then the smallest angle is :
Which of the following is correct?
Codes:
1 2 3- There are 40 persons in a palace. If every person shakes hands with every other person, what will be the total number of handshakes?
- How many words can be formed with or without meaning by using all letters of the word DELHI?
A rectangular wall is divided into four squares of equal size where there are two rows each having two squares. The top left square is coloured with green. If, including green, there are three colours available and each square is coloured using any one of these three colours such that no two adjacent squares get painted with the same colour; then how many colour combinations are possible ?
