Step-by-step Solution:

Statement I: The number of different numbers each of 6 digits that can be formed by using all the digits 1, 2, 1, 0, 2, 2 is 50. We need to calculate the number of distinct 6-digit numbers that can be formed using the digits 1, 2, 1, 0, 2, 2. Note that 0 cannot be the first digit in a 6-digit number. 1. Total digits: 6 (with repetitions: two 1's and three 2's, and one 0). 2. Total permutations without restrictions: \[ \frac{6!}{2! \times 3!} = \frac{720}{2 \times 6} = 60 \] 3. Subtract the permutations where 0 is the first digit: - Fix 0 as the first digit. - Remaining digits: 1, 2, 1, 2, 2. - Permutations of the remaining digits: \[ \frac{5!}{2! \times 3!} = \frac{120}{2 \times 6} = 10 \] 4. Valid 6-digit numbers: \[ 60 - 10 = 50 \] \[Statement I is true.\] Statement II: There are 4536 possibilities of writing the four-digit numbers which have all distinct digits. We need to calculate the number of 4-digit numbers with all distinct digits. 1. First digit (thousands place): Cannot be 0, so there are 9 possible choices (1-9). 2. Second digit (hundreds place): Can be 0, but must be distinct from the first digit, so 9 choices. 3. Third digit (tens place): Must be distinct from the first two digits, so 8 choices. 4. Fourth digit (units place): Must be distinct from the first three digits, so 7 choices. Total number of 4-digit numbers with all distinct digits: \[ 9 \times 9 \times 8 \times 7 = 4536 \] Statement II is true. Conclusion: - Statement I is true. - Statement II is true. Therefore, the correct answer is: A. Both Statement I and Statement II are true