Question

A four digit number is to be formed using the digits \(0,1,3,5\) and \(7 .\) How many of them are even numbers? (Each digit can occur for only one time). (1) 48 (2) 60 (3) 24 (4) 120

Short answer

Answer: There are 14 even four-digit numbers that can be formed using the digits {0, 1, 3, 5, 7}.

Step-by-step solution

Determine the even ones digit

Since the only even number in our given set {\(0,1,3,5,7\)} is \(0\), any even four-digit number formed with these digits will end in \(0\).

Count the permutations

We now need to count how many even four-digit numbers can be formed using these digits. Since the even number ends in \(0\), there are only 3 remaining places to fill. For the first digit, there are 3 choices( because we can't use the digit 0 at the first place) - {\(1,3,5,7\)} For the second digit, now there are 3 choices remaining (since 0 is used and one number from the above used) For the third digit, there are 2 choices remaining. Finally, the last digit is already decided by the even digit \(0\) Now, we can take the product of the number of choices for each of the remaining places to calculate the total number of possible permutations. Total permutations = \(3 \times 3 \times 2 = 18\) However, these permutations do not account for the cases in which the first digit is \(0\). Since \(0\) cannot be the first digit of the four-digit number, we must subtract these cases. There are \(2 \times 2 = 4\) such cases (choosing from 3,5,7 for the second and third places) So, number of valid 4 digit even numbers formed using the given digits = Total Permutations - Invalid Permutations = \(18 - 4 = 14\) Therefore, among the given options, none of them are correct. There are 14 even four-digit numbers that can be formed using the digits {\(0,1,3,5,7\)}.

Key concepts

Permutations

In combinatorics, permutations are the ways to arrange elements in a specific order. When forming four-digit numbers, each digit's position matters. For example, in numbers such as 1537 and 1753, the order of digits is different, making each a unique permutation.

To find permutations, we consider both the available elements and the positions we have to fill. When all elements are distinct, the number of permutations of length \( n \) from a set of elements is given by \( n! \). Permutations are crucial for calculating how we can arrange the remaining digits after choosing one digit for the units place to ensure the number is even.

Even Numbers

An even number is a number that is divisible by 2 without any remainder. In the context of permutations, even numbers can be identified by the digit in the units place. That digit must be even.

In this problem, the digits available are \( \{0, 1, 3, 5, 7\} \), and only \( 0 \) is even. Thus, any even four-digit number made with these digits must have \( 0 \) in its last position. This is because only decimals ending with \( 0 \) are divisible by \( 2 \).

Four-Digit Numbers

A four-digit number consists of a thousands, hundreds, tens, and units digit. It is important to remember that in standard notation, the number cannot start with zero, as then it would not be a true four-digit number.

The hundreds, tens, and units positions must be filled carefully after ensuring that the leading digit is not \( 0 \). In this scenario, because \( 0 \) is in the units position, only the remaining digits \( 1, 3, 5, \) and \( 7 \) are considered for the first three positions. One digit must also be excluded from the start due to the digit being placed in the units towards the end, thus creating a valid sequence of four figures.

Digit Restrictions

Digit restrictions occur when certain criteria limit how or where digits can be placed. In the problem, one such restriction is that \( 0 \) cannot be the first digit. Another restriction is that each digit can only appear once in a number, so no repetition is allowed.

These constraints ensure that all four-digit sequences generated are valid, meaning they do not violate the rules set by numbering conventions or problem specifications. Digit restrictions help in narrowing down permissible arrangements significantly and thus simplify permutation calculations.

• You need to exclude \( 0 \) from the thousands place.
• Each permutation must utilize the remaining digits without repetition.