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 __________.

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Step 1: Understand the set $ S $ - $ S $ is the set of all seven-digit numbers formed using the digits 0, 1, and 2. - A seven-digit number has the form $ d_1 d_2 d_3 d_4 d_5 d_6 d_7 $, where each $ d_i \in \{0, 1, 2\} $. - Since it’s a seven-digit number, the first digit $ d_1 $ cannot be 0 (otherwise, it wouldn’t be a seven-digit number; e.g., 0210222 would be interpreted as 210222, a six-digit number). - Thus: \qquad - $ d_1 \in \{1, 2\} $ (2 choices), \qquad - $ d_2, d_3, \ldots, d_7 \in \{0, 1, 2\} $ (3 choices each). - Total number of seven-digit numbers in $ S $: $$ 2 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 = 2 \times 3^6. $$ - Compute $ 3^6 = 729 $, so: $$ 2 \times 729 = 1458. $$ - Therefore, $ |S| = 1458 $. The example confirms this: - 2210222 is a seven-digit number with digits from \{0, 1, 2\}, so it’s in $ S $. - 0210222 is not a seven-digit number (it’s 210222, a six-digit number), so it’s not in $ S $. Step 2: Define the condition - We need to find the number of elements $ x \in S $ such that the number of digits 0 and the number of digits 1 in $ x $ are equal. - Let: \qquad - Number of 0s in $ x = k $, \qquad - Number of 1s in $ x = k $, \qquad - Number of 2s in $ x = 7 - (k + k) = 7 - 2k $. - Since the total number of digits is 7, and the number of 2s must be non-negative: $$ 7 - 2k \geq 0 \implies 2k \leq 7 \implies k \leq 3.5 \implies k \leq 3 \text{ (since } k \text{ is an integer)}. $$ - Also, $ k \geq 0 $. So, $ k $ can be 0, 1, 2, or 3. Step 3: Count the numbers for each $ k $ For each value of $ k $, compute the number of valid seven-digit numbers, considering the restriction on the first digit. Case 1: $ k = 0 $ - 0s: 0, 1s: 0, 2s: $ 7 - 2 \times 0 = 7 $. - All digits are 2: the number is 2222222. - First digit is 2, which is fine. - Number of such numbers: 1. Case 2: $ k = 1 $ - 0s: 1, 1s: 1, 2s: $ 7 - 2 \times 1 = 5 $. - Total digits: 7. - Choose 1 position out of 7 for the 0: $ \binom{7}{1} = 7 $. - From the remaining 6 positions, choose 1 for the 1: $ \binom{6}{1} = 6 $. - The remaining 5 positions are 2s. - Total ways (without considering the first digit): $ 7 \times 6 = 42 $. - Now, exclude numbers where the first digit is 0: \qquad - First digit is 0: Fix position 1 as 0 (1 way). \qquad - Choose 1 position out of the remaining 6 for the 1: $ \binom{6}{1} = 6 $. \qquad - Remaining 5 positions are 2s. \qquad - Number of invalid cases: 6. - Valid cases: $ 42 - 6 = 36 $. Case 3: $ k = 2 $ - 0s: 2, 1s: 2, 2s: $ 7 - 2 \times 2 = 3 $. - Choose 2 positions out of 7 for the 0s: $ \binom{7}{2} = \frac{7 \times 6}{2} = 21 $. - From the remaining 5 positions, choose 2 for the 1s: $ \binom{5}{2} = \frac{5 \times 4}{2} = 10 $. - Remaining 3 positions are 2s. - Total ways: $ 21 \times 10 = 210 $. - Exclude cases where the first digit is 0: \qquad - First digit is 0: Fix position 1 as 0. \qquad - Choose 1 more position out of positions 2 to 7 for the other 0: $ \binom{6}{1} = 6 $. \qquad - From the remaining 5 positions, choose 2 for the 1s: $ \binom{5}{2} = 10 $. \qquad - Remaining 3 positions are 2s. \qquad - Invalid cases: $ 6 \times 10 = 60 $. - Valid cases: $ 210 - 60 = 150 $. Case 4: $ k = 3 $ - 0s: 3, 1s: 3, 2s: $ 7 - 2 \times 3 = 1 $. - Choose 3 positions out of 7 for the 0s: $ \binom{7}{3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35 $. - From the remaining 4 positions, choose 3 for the 1s: $ \binom{4}{3} = 4 $. - Remaining 1 position is a 2. - Total ways: $ 35 \times 4 = 140 $. - Exclude cases where the first digit is 0: \qquad - First digit is 0: Fix position 1 as 0. \qquad - Choose 2 more positions out of positions 2 to 7 for the other 0s: $ \binom{6}{2} = 15 $. \qquad - From the remaining 4 positions, choose 3 for the 1s: $ \binom{4}{3} = 4 $. \qquad - Remaining position is a 2. \qquad - Invalid cases: $ 15 \times 4 = 60 $. - Valid cases: $ 140 - 60 = 80 $. Step 5: Sum the valid cases - $ k = 0 $: 1, - $ k = 1 $: 36, - $ k = 2 $: 150, - $ k = 3 $: 80. - Total: $$ 1 + 36 + 150 + 80 = 267. $$ Final Answer: The number of elements $ x \in S $ where the number of 0s equals the number of 1s is $ \boxed{267} $.

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Correct Answer: 267

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