Question

(i) Find the number of all even 5-digit positive integers written with 1, 2, 3, 4, 5 such that none of the digits is repeated?
(ii) Find the number of all even 6-digit positive integers written with 0, 1, 2, 3, 4, 5 such that none of the digits is repeated?

Answer 1

Hint: We start solving the problem by recalling the fact that the digit in the unit place should be even in order to get an even number. So, we check the total number of possibilities for the unit place first and then find the total number of arrangements for the remaining places in the number using the fact that the permutation of n distinct objects taken all at a time without repetition is $n!$ ways. We also check that the total no. of possibilities for the first digit as we need only 6 digit even positive numbers in problem (2).

Complete step-by-step solution:
(i) According to the problem, we need to find the total number of even 5-digit positive integers written with 1, 2, 3, 4, 5 such that none of the digits is repeated.
Now, we need to find the total no. of possibilities for the place of given numbers.
We know that the digit in the unit place of any even positive number should be even. From the given digits we have only two even numbers i.e., 2 and 4. So, we can only fill the unit place with one of these numbers. So, we have 2 possibilities for filling the unit place to get an even number.
Now, we need to assign numbers for the remaining 4 places with the remaining four digits as it is given that none of the digits should be repeated. We need to arrange 4 places with the remaining four distinct numbers as one of the digits is already used for unit place. We can see that this is similar to permutation of n distinct objects taken all at a time without repetition i.e., $n!$ ways.
So, we can fill the remaining 4 places in $4!$ ways. We get the total number of even 5-digit positive integers as $4!\times 2$.
We know that $n!=n\times \left( n-1 \right)\times ......\times 2\times 1$. So, we get $\left( 4\times 3\times 2\times 1 \right)\times 2=48$ numbers.
We get the total number of even 5-digit positive integers with 1, 2, 3, 4, and 5 as 48.
(ii) According to the problem, we need to find the total number of even 6-digit positive integers written with 0,1, 2, 3, 4, 5 such that none of the digits is repeated.
Now, we need to find the total no. of possibilities for the place of given numbers.
We know that the digit in the unit place of any even positive number should be even. From the given digits we have only three even numbers i.e., 0, 2, and 4. So, we can only fill the unit place with one of these numbers.
If we assign unit place with 0, then we need to assign numbers for the remaining 5 places with the remaining 5 digits as it is given that none of the digits should be repeated. We need to arrange 5 places with the remaining five distinct numbers as zero is already used for unit place. We can see that this is similar to permutation of n distinct objects taken all at a time without repetition i.e., $n!$ ways.
So, we can fill the remaining 5 places in $5!$ ways. We get the total number of even 6-digit positive integers having unit place arranged with 0 is $5!$.
If we assign unit place with 2 or 4, then we have 2 possibilities of filling the unit place to get an even number. We need to assign numbers for the remaining 5 places with the remaining 5 digits as it is given that none of the digits should be repeated. But the first place cannot be assigned with 0 as it will make the obtained number as a 5-digit number. So, we need to assign a number other than 0 in first which makes us having 4 possibilities of filling first place as 0 is not accounted for first place from the remaining 5 digits. Now, we need to arrange the remaining 4 places with the remaining four distinct numbers as zero is already used for unit place. We can see that this is similar to permutation of n distinct objects taken all at a time without repetition i.e., $n!$ ways.
So, we can fill the remaining 4 places in $4!$ ways. We get the total number of even 6-digit positive integers having unit place arranged with 2 or 4 is $4\times 2\times 4!=8\times 4!$.
Do, we get the total number of even 6-digit positive integers as $5!+\left( 8\times 4! \right)$.
We know that $n!=n\times \left( n-1 \right)\times ......\times 2\times 1$. So, we get $\left( 5\times 4\times 3\times 2\times 1 \right)+\left( 4\times 3\times 2\times 1 \right)\times 8=120+192=312$ numbers.
We get the total number of even 6-digit positive integers with 0, 1, 2, 3, 4, and 5 as 312.

Note: We should solve the problem (ii) like we solve the problem (i) as we have the digit 0 here which should not be placed in the first place of the number to get a 6-digit even positive number. We should make sure about the information given in the problem with the repetition of the digits in the required numbers while solving the problem. If the digits are repeated then we cannot use the formula of permutation we just used.