Question

How many even numbers greater than 300 can be formed with the digits 1,2,3,4,5 if repetition of digits in a number is not allowed?

Answer 1

First we will find the five digit numbers formed by the five digits available with a condition that the last digit is 2 or 4. Then, we will find the four digit numbers formed by the five digits available with a condition that the last digit is 2 or 4. At last we will find the three digit numbers formed by the five digits available with a condition that the last digit is 2 or 4. In three digits we have to omit all the numbers formed starting with 1 or 2. Then we add all the numbers available.

Complete step-by-step answer:
If the number is > 300 if must have at least 3 digits
There are only 5 digits to choose so it must have either 3 4 or 5 digits.
Number can be 3, 4, or 5 digit (4 and 5 digit numbers will obviously be more than 300).
5 digit even numbers:
The last digit has to be 2 or 4. Last digit again must be either 2 or 4 (for the number to be even), so 2 choices.
First 4 digits can be arranged in 4! of ways.
So total ways of 5 digit even numbers possible is$4!{\times }2 = 48$;
4 digit even numbers:
The last digit has to be 2 or 4. Last digit again must be either 2 or 4 (for the number to be even), so 2 choices.
The Way to choose the 3 digits out of 4 digits left when order of the digits matters is${}^4{P_3}$.
So total ways of 4 digit even numbers possible is ${}^4{P_3}{\times }2 = 48$;
3 digit even numbers:
The last digit has to be 2 or 4. Last digit again must be either 2 or 4 (for the number to be even), so 2 choices.
The Way to choose 2 digits out of 4 digits left when order of the digits matters is ${}^4{P_2}$.
So total ways of 3 digit even numbers possible is ${}^4{P_2}{\times}2 = 24$.
But out of these 24 3-digit even numbers some will be less than 300, the ones starting with 1 or 2. There are 9 such numbers
Numbers starting with 1, second place = Ways to choose rest 1 digit out of 3 digits left when order of the digits matters is ${}^3{P_1}$=3.
The last digit has to be 2 or 4. Last digit again must be either 2 or 4 (for the number to be even), so 2 choices.
So the total number of Numbers starting with 1, in a 3 digit even numbers possible is ${}^3{P_1}{\times}2 = 6$.

Numbers starting with 2 second place = Ways to choose rest 1 digits out of 4 digits left when order of the digits matters is ${}^3{P_1}$=3.
The last digit has to be 4. so 1 choice.
So the total ways of Numbers starting with 2, in a 3 digit even number possible is ${}^3{P_1}{\times}1 = 3$.
which means that ways of 3 digit even numbers more than 300 is $24 - 9 = 15$;
So total $48 + 48 + 15 = 111$.

111 even numbers greater than 300 can be formed with the digits 1,2,3,4,5 if repetition of digits in a number is not allowed.

Note: In questions like these, we have two categories in 3 digit numbers: Greater than 300 and less than 300. And we need to see the options separately of both the types.
Thus, in this case, we will write down all the possible ways of having less than 300 separately and then minus from the total number to get greater than 300 numbers