We are asked to form four-digit even numbers using the digits 0, 1, 2, and 3 without repetition.
- The first digit cannot be 0 (since it would not be a four-digit number), so the first digit can be 1, 2, or 3. This gives us 3 choices for the first digit.
- The last digit must be even, so the last digit can be 0 or 2. Therefore, we have 2 choices for the last digit.
- After selecting the first and last digits, we are left with 2 digits from the remaining available digits. So, for the second digit, we have 2 choices, and for the third digit, we have 1 choice.
Thus, the total number of possible four-digit even numbers is:
3 × 2 × 2 × 1 = 6
So, the correct answer is 6.
Match List-I with List-II
List-I | List-II |
---|---|
(A) \(^{8}P_{3} - ^{10}C_{3}\) | (I) 6 |
(B) \(^{8}P_{5}\) | (II) 21 |
(C) \(^{n}P_{4} = 360,\) then find \(n\). | (III) 216 |
(D) \(^{n}C_{2} = 210,\) find \(n\). | (IV) 6720 |
Choose the correct answer from the options given below: