Question

The number of even numbers greater than 100 that can be formed by the digits 0, 1, 2, 3 (no digit being repeated) is
A. 20
B. 30
C. 40
D. None of these

Answer 1

Hint: In order to solve this question, we should know that greater than 100 means all the 3-digit numbers and 4-digit numbers. Also, we should be aware that even numbers either end with 0 or 2. So, we will consider each of the possible cases like, 3-digit number ending with 2 and 0, 4-digit number ending with 2 and 0 from the given digits. And then we will be able to calculate all the numbers of even numbers greater than 100 that are formed by 0, 1, 2, 3.

Complete step-by-step answer:
In this question, we have been asked to find the number of even numbers greater than 100 that can be formed by the digits 0, 1, 2, 3 such that no digits are repeated. To solve this question, we should know that all 3-digit and 4-digit numbers are greater than 100. Also, that even numbers will either end with 0 or 2 as per the given digits. So, we get 4 possible cases to find our answer. They are, 3-digit numbers ending with 0, 3-digit number ending with 2, 4-digit number ending with 0 and 4-digit number ending with 2. Therefore, to get the answer, we will calculate the number of numbers that can be included in each category.
Case I - 3-digit number ending with 0.
Now, as the last digit is 0, we can say that none of the other digits would be 0 as digits cannot be repeated. So, we can say that the 1st digit can be any one out of 1, 2, 3, that is 3 possible choices and one of these 3 would be chosen. So, we will be left with 2 choices for the 2nd digit. Hence, we can say that the number of 3-digit numbers ending with 0 are $3\times 2=6$.
Case II - 3-digit number ending with 2.
Now, as the last digit is 2, we can say that none of the other digits would be 2 as digits cannot be repeated. We know that 0 cannot be the 1st digit of a number. So, we can say that we have only 1, 3, that is 2 possible choices for the 1st digit. Now, out of these 2 one would be selected for 1st digit and we would be left with only one choice for the 2nd digit. But we can put 0 in the 2nd digit, so we have 2 choices for the 2nd digit. Hence, we can say that the number of 3-digit numbers ending with 2 are $2\times 2=4$.
Case III - 4-digit number ending with 0.
Now, as the last digit is 0, we can say that none of the other digits would be 0. So, for the 1st digit, we have 3 choices, that is 1, 2, 3. And for the 2nd digit, we would then have 2 choices left as 1 digit is already chosen for the 1st digit. And there will be only 1 choice left for the third digit. So, the number of 4-digit numbers ending with 0 are $3\times 2\times 1=6$.
 Case IV - 4-digit number ending with 2.
Now, as the last digit is 2, we can say that none of the other digits would be 2. We know that 0 cannot be the 1st digit of a number. So, we can say that we have only 1, 3, as the 2 possible choices for the 1st digit. So, we would be left with one choice out of 1, 3 for the 2nd digit along with 0, which can be put in the 2nd digit. So, we have 2 choices for the 2nd digit. And so, we would be left with only 1 choice for the third digit. Hence, we can say that the number of 4-digit numbers ending with 2 are $2\times 2\times 1=4$.
Now, to calculate the total number of even numbers greater than 100, formed by the digits 0, 1, 2, 3, it would be the sum of the numbers formed in case I, II, III, IV. So, we get the number of possible even numbers as 6 + 4 + 6 + 4 = 20.
Hence, the correct answer is option A.

Note: There is a possibility, while solving this question that we do not include the 3-digit numbers by putting 0 in the first place of 4-digit numbers. So, it is better to consider each of the cases separately to reach the final answer.