Find the number of all even 5-digit positive integers written with 1, 2, 3, 4, 5 such that none of its digit is repeated.

Hint: Now we know that 5 – digit even number will have an even number in its units place. So first we will fix 2 in units place and then find the number of digits formed in this case. This can be done as the number of ways of arranging n objects in n places is n! Now similarly we will find the number of 5 – digit numbers formed when we fix 4 in units place. Now we will add all the possibilities to get the total number of 5 – digit even number possible.

Complete step-by-step answer:
Now we are given with the digits 1, 2, 3, 4, 5.
We want to form a 5 digit even integer from these digits. We know if a number is even then there must be an even digit in the units place. Now among 1, 2, 3, 4, 5. We just have 2 even digits which are 2 and 4.
Now let us form a number with 2 in its units place.
Now we have fixed the units place of 5 digit numbers and hence now we have to arrange the rest 4 digits in 4 places without repetition.
Number of arrangements of n objects in n places without repetition is given by n!
Hence 4 digits can be arranged in 4 places in 4! ways
Hence we get, the number of even numbers formed with 2 in its units place is
4! = 24. ……………………… (1)
Now let us form a number with 4 in its units place.
Now again we have fixed the units place of 5 digit numbers and hence now we just have to arrange the rest 4 digits in 4 places without repetition.
Number of arrangements of n objects in n places without repetition is given by n!
Hence rest 4 digits can be arranged in 4 places in 4! ways
Hence we get, the number of even numbers formed with 2 in its units place is
4! = 24 ……………………….. (2).
Hence from (1) and (2) we get,
Number of even numbers formed = 24 + 24 = 48
Hence the number of 5 digit even numbers formed by digits 1, 2, 3, 4 and 5 is 48.

Note: Now we can also solve this problem directly. The number of possible options for units placed is 2. Now after fixing units place we are left with 4 digits, hence we arrange them in remaining 4 places in 4! Ways. Hence the total number of 5 – digit numbers formed = 2 × 4! = 48.