Erin is thinking of a 3-digit number.It uses the digits 1,7 and 4.How many numbers can I make that are odd? How many numbers can I make that are even ?Thanks!
Oh, I love numbers and I love odd and even ones too! Let's explore Erin's 3-digit number using the digits 1, 7, and 4.
To make an odd number, we need to make sure the last digit is an odd number. So, how many odd numbers can we make with these digits? Hmm... well, the only odd digit we have is 1. Therefore, there can only be one odd number, which is 171.
Now, let's move on to making even numbers. For a number to be even, its last digit must be even as well. In this case, we have the digit 4, which is indeed an even number. So, we can form two even numbers, which are 174 and 147.
So to summarize, you can make one odd number (171) and two even numbers (174 and 147). Hope that puts a smile on your face!
To determine the number of odd and even numbers that can be made with the digits 1, 7, and 4, we need to consider the possible combinations.
First, we establish the restrictions:
1. The number must be 3 digits.
2. The digits used are 1, 7, and 4.
Odd numbers:
To create an odd number, the last digit must be either 1 or 7, as these are the only odd digits provided.
1. We have two options for the last digit (1 or 7).
2. For the first two digits, we can arrange the remaining two digits (1 and 4) in any order.
Using the permutation formula nPr (n factorial divided by (n-r) factorial), we can calculate the number of odd numbers:
Number of odd numbers = 2 x 2P2 = 2 x (2!/0!) = 2 x 2 = 4
Even numbers:
To create an even number, the last digit must be 4 since it is the only even digit provided.
1. We have one option for the last digit, which is 4.
2. For the first two digits, we can arrange the remaining two digits (1 and 7) in any order.
Using the same permutation formula, we can calculate the number of even numbers:
Number of even numbers = 1 x 2P2 = 1 x (2!/0!) = 1 x 2 = 2
In conclusion:
The number of odd numbers that can be made is 4.
The number of even numbers that can be made is 2.
To determine how many odd and even numbers can be formed using the digits 1, 7, and 4, we need to consider the possible arrangements of these digits.
First, let's determine the number of possible 3-digit numbers that can be formed using these digits. Since repetition is allowed, we can use the formula for permutations: n! / (n1! * n2! * ... * nk!), where n is the total number of objects and n1, n2, etc. represent the number of times each object may appear.
In this case, n = 3 since there are three digits (1, 7, and 4), and n1 = n2 = n3 = 1 since each digit can only appear once. Therefore, the total number of possible 3-digit numbers is:
3! / (1! * 1! * 1!) = 3! / 1 = 3 × 2 × 1 / 1 = 6
Now, let's determine the number of possible odd numbers. For a number to be odd, its last digit must be 1 or 7. Therefore, we have two choices for the last digit and two remaining digits (1 and 7) for the first two digits, resulting in a total of 2 × 2 = 4 possible odd numbers.
Next, let's determine the number of possible even numbers. For a number to be even, its last digit must be 4. Therefore, we only have one choice for the last digit, and there are two remaining digits (1 and 7) for the first two digits. This gives us a total of 1 × 2 = 2 possible even numbers.
So, to answer your questions:
- The number of 3-digit numbers that can be formed using 1, 7, and 4 is 6.
- The number of odd numbers that can be formed is 4.
- The number of even numbers that can be formed is 2.