
All possible four digit numbers are formed using the digits $0,1,2,3$ so that no number has repeated digits. The number of even numbers among them is?
$
\left( 1 \right)9 \\
\left( 2 \right)18 \\
\left( 3 \right)10 \\
$
$\left( 4 \right)$ None of these
Answer
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Hint: First we need to figure out what are the possible even digits from the given set of digits. Then we have to consider each of the individual even digits and calculate the number of even numbers that can be formed keeping the even digit at the last place. An integer is said to be even if dividing the number by $2$ does not give any remainder.
Complete step-by-step answer:
It is given that four-digit numbers are formed by using the digits without repetition.
We need to find the number of even numbers that can be formed using these digits. There are only two even numbers i.e.$0$ and $2$.
So we have two possibilities for the number being even.
First case: If the last digit of the four-digit number is $0$ then we are left with three choices for the first place in the number, two choices for the second place, and one choice for the third place.
Therefore, the required number of ways to get an even number where the last digit is $0$ is
$
= 3 \times 2 \times 1 \\
= 6 \\
$
Second case: If the last digit of the four-digit number is $2$ then we are left with two choices for the first place in the number because a number cannot start with $0$ so we need to exclude $0$ from the remaining three choices. Next, we have two choices for the second place i.e after filling the first place we are left with $0$ and another number. Lastly, we have one choice for the third place after the second place is filled.
Therefore, the required number of ways to get an even number where the last digit is $2$ is
\[
= 2 \times 2 \times 1 \\
= 4 \\
\]
$\therefore $ By using these four digits, the total number of even numbers formed is :
$
= 6 + 4 \\
= 10 \\
$
Hence, the correct option is $\left( 3 \right)10$.
So, the correct answer is “Option (3)”.
Note: One must take note of the fact that we cannot start a number with $0$. This has to be kept in mind while calculating the number of different even numbers that can be generated. One must also have prior knowledge of what an even number is and should be able to recognize them from the given digits.
Complete step-by-step answer:
It is given that four-digit numbers are formed by using the digits without repetition.
We need to find the number of even numbers that can be formed using these digits. There are only two even numbers i.e.$0$ and $2$.
So we have two possibilities for the number being even.
First case: If the last digit of the four-digit number is $0$ then we are left with three choices for the first place in the number, two choices for the second place, and one choice for the third place.
Therefore, the required number of ways to get an even number where the last digit is $0$ is
$
= 3 \times 2 \times 1 \\
= 6 \\
$
Second case: If the last digit of the four-digit number is $2$ then we are left with two choices for the first place in the number because a number cannot start with $0$ so we need to exclude $0$ from the remaining three choices. Next, we have two choices for the second place i.e after filling the first place we are left with $0$ and another number. Lastly, we have one choice for the third place after the second place is filled.
Therefore, the required number of ways to get an even number where the last digit is $2$ is
\[
= 2 \times 2 \times 1 \\
= 4 \\
\]
$\therefore $ By using these four digits, the total number of even numbers formed is :
$
= 6 + 4 \\
= 10 \\
$
Hence, the correct option is $\left( 3 \right)10$.
So, the correct answer is “Option (3)”.
Note: One must take note of the fact that we cannot start a number with $0$. This has to be kept in mind while calculating the number of different even numbers that can be generated. One must also have prior knowledge of what an even number is and should be able to recognize them from the given digits.
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