
How many four-digit even number can be formed using the digits $0$,$1$, $2$, $3$ and $4$. If repetition of the digits is not permitted?
Answer
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Hint: For a number to be $4$ - digited and even it must not start with $0$ because if it starts with $0$ then it will become $3$ digited number, not a $4$ digited number and it must end with the digits $0$, $2$, $4$ to be called it as even. Now here we have two cases, so we will calculate the number of ways to each case and add them to get the result.
Complete step by step answer:
Given digits $0$,$1$, $2$, $3$ and $4$.
The number of digits is $5$.
Case – 1: Number having unit digit $0$.
Let $-$be the four empty dashes where a digit can be accommodated. When the unit digit is $0$, there can be $5-1=4$ digits left to be assigned to the rest of the three spaces. Provided that repetition is not allowed, the thousandth place can be hold by any $4$ digits from $\left( 1,2,3,4 \right)$ after which we have $4-1=3$ digits remaining so the hundredth place can be filled with any of the $3$ digits and likewise the tenth place can be filled with any of the $2$ digits.
$\therefore $ The total number of ways $=4\times 3\times 2=24$
Case – 2: Number ending with $2$ or $4$.
If the unit digit of the number is $2$ then we have $5-1=4$ digits left $\left( 0,1,3,4 \right)$. The thousandth digit of the number can be filled in $3$ ways $\left( 1,3,4 \right)$ but it cannot hold $0$ else it will generate a $3$-digit number. The hundredth digit of the number can be filled in $3$ ways (including $0$ and any other digit left). And similarly, the tenth digit of the number can be filled in $2$ ways (including $0$ and any other digit left)
So, the total number of numbers will be $3\times 3\times 2=18$.
Now the same case is applicable for the case in which the number will end with $4$. So, the total number of ways in the case – 2 is $2\times 18=36$.
$\therefore $ The total four-digit even numbers formed are $24+36=60$.
Note:
Students should be careful about repetition. In the problem, they have mentioned that repetition is not allowed, so we solve the problem in this manner. But when they have mentioned that repetition is allowed, there should be a difference in the result.
Complete step by step answer:
Given digits $0$,$1$, $2$, $3$ and $4$.
The number of digits is $5$.
Case – 1: Number having unit digit $0$.
Let $-$be the four empty dashes where a digit can be accommodated. When the unit digit is $0$, there can be $5-1=4$ digits left to be assigned to the rest of the three spaces. Provided that repetition is not allowed, the thousandth place can be hold by any $4$ digits from $\left( 1,2,3,4 \right)$ after which we have $4-1=3$ digits remaining so the hundredth place can be filled with any of the $3$ digits and likewise the tenth place can be filled with any of the $2$ digits.
$\therefore $ The total number of ways $=4\times 3\times 2=24$
Case – 2: Number ending with $2$ or $4$.
If the unit digit of the number is $2$ then we have $5-1=4$ digits left $\left( 0,1,3,4 \right)$. The thousandth digit of the number can be filled in $3$ ways $\left( 1,3,4 \right)$ but it cannot hold $0$ else it will generate a $3$-digit number. The hundredth digit of the number can be filled in $3$ ways (including $0$ and any other digit left). And similarly, the tenth digit of the number can be filled in $2$ ways (including $0$ and any other digit left)
So, the total number of numbers will be $3\times 3\times 2=18$.
Now the same case is applicable for the case in which the number will end with $4$. So, the total number of ways in the case – 2 is $2\times 18=36$.
$\therefore $ The total four-digit even numbers formed are $24+36=60$.
Note:
Students should be careful about repetition. In the problem, they have mentioned that repetition is not allowed, so we solve the problem in this manner. But when they have mentioned that repetition is allowed, there should be a difference in the result.
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