Question

Let S be the set of all seven-digit numbers that can be formed using the digits 0, 1 and 2. For example, 2210222 is in S, but 0210222 is NOT in S. Then the number of elements x in S such that at least one of the digits 0 and 1 appears exactly twice in x, is equal to _____________.

Answer 1

762

Answer 2

Let S be the set of all seven-digit numbers that can be formed using the digits 0, 1 and 2. A seven-digit number must have the first digit non-zero. The digits allowed are 0, 1, 2. So, the first digit can be 1 or 2. The remaining six digits can be 0, 1, or 2. The total number of elements in S is $2 \times 3^6 = 2 \times 729 = 1458$.

We are looking for the number of elements x in S such that at least one of the digits 0 and 1 appears exactly twice in x. Let A be the set of numbers in S where the digit 0 appears exactly twice. Let B be the set of numbers in S where the digit 1 appears exactly twice. We want to find $|A \cup B| = |A| + |B| - |A \cap B|$.

To find $|A|$, we count the number of 7-digit numbers in S with exactly two 0s. The digits used are two 0s, and five digits from {1, 2}. The first digit cannot be 0. Case 1: The first digit is 1. The first digit is 1. We need to place two 0s and four digits from {1, 2} in the remaining 6 positions. The number of ways to choose the positions for the two 0s from the remaining 6 positions is $\binom{6}{2}$. The remaining 4 positions must be filled with digits from {1, 2}. There are $2^4$ ways. Number of elements in this case = $\binom{6}{2} \times 2^4 = 15 \times 16 = 240$.

Case 2: The first digit is 2. The first digit is 2. We need to place two 0s and four digits from {1, 2} in the remaining 6 positions. The number of ways to choose the positions for the two 0s from the remaining 6 positions is $\binom{6}{2}$. The remaining 4 positions must be filled with digits from {1, 2}. There are $2^4$ ways. Number of elements in this case = $\binom{6}{2} \times 2^4 = 15 \times 16 = 240$.

Total number of elements in S with exactly two 0s is $|A| = 240 + 240 = 480$.

To find $|B|$, we count the number of 7-digit numbers in S with exactly two 1s. The digits used are two 1s, and five digits from {0, 2}. The first digit cannot be 0. Case 1: The first digit is 1. The first digit is 1. We need to place one more 1 and five digits from {0, 2} in the remaining 6 positions. The number of ways to choose the position for the second 1 from the remaining 6 positions is $\binom{6}{1}$. The remaining 5 positions must be filled with digits from {0, 2}. There are $2^5$ ways. Number of elements in this case = $\binom{6}{1} \times 2^5 = 6 \times 32 = 192$.

Case 2: The first digit is 2. The first digit is 2. We need to place two 1s and four digits from {0, 2} in the remaining 6 positions. The number of ways to choose the positions for the two 1s from the remaining 6 positions is $\binom{6}{2}$. The remaining 4 positions must be filled with digits from {0, 2}. There are $2^4$ ways. Number of elements in this case = $\binom{6}{2} \times 2^4 = 15 \times 16 = 240$.

Total number of elements in S with exactly two 1s is $|B| = 192 + 240 = 432$.

To find $|A \cap B|$, we count the number of 7-digit numbers in S with exactly two 0s and exactly two 1s. The digits used are two 0s, two 1s, and three 2s (since $2+2+3=7$). The first digit cannot be 0. Total permutations of two 0s, two 1s, and three 2s is $\frac{7!}{2! 2! 3!} = \frac{5040}{2 \times 2 \times 6} = \frac{5040}{24} = 210$. We subtract the number of such arrangements that start with 0. If the first digit is 0, the remaining 6 digits must contain one 0, two 1s, and three 2s. The number of permutations of these 6 digits is $\frac{6!}{1! 2! 3!} = \frac{720}{1 \times 2 \times 6} = \frac{720}{12} = 60$. Number of elements in S with exactly two 0s and exactly two 1s is $|A \cap B| = 210 - 60 = 150$.

Using the principle of inclusion-exclusion: $|A \cup B| = |A| + |B| - |A \cap B| = 480 + 432 - 150 = 912 - 150 = 762$.

The number of elements x in S such that at least one of the digits 0 and 1 appears exactly twice in x is 762.

The final answer is $\boxed{762}$.