The sum of all the four-digit numbers that can be formed using all the digits 2, 1, 2, 3 is equal to

To solve this problem, we need to understand that each digit will appear in each position (thousands, hundreds, tens, ones) the same number of times.

Step 1: Calculate the total number of four-digit numbers that can be formed using the digits 2, 1, 2, 3. Since there are 4 digits but one of them is repeated, we use the formula for permutations of a multiset: 4! / 2! = 12.

Step 2: Calculate the sum for each position. Each digit will appear in each position 12 / 4 = 3 times. So, the sum for each position is (2+1+2+3) * 3 = 24.

Step 3: Calculate the total sum. The sum of all the numbers is the sum for each position times the value of the position (thousands, hundreds, tens, ones). So, the total sum is 24 * (1000 + 100 + 10 + 1) = 24 * 1111 = 26664.

So, the sum of all the four-digit numbers that can be formed using all the digits 2, 1, 2, 3 is 26664.

To solve this problem, we need to understand that each digit will appear in each position (thousands, hundreds, tens, ones) the same number of times.

Step 1: Calculate the total number of four-digit numbers that can be formed with the digits 2, 1, 2, 3. Since we have one repeated digit (2), the total number of permutations is 4! / 2! = 12.

Step 2: Calculate the sum of the digits 2, 1, 2, 3 which is 2 + 1 + 2 + 3 = 8.

Step 3: Each digit will appear in each position (thousands, hundreds, tens, ones) the same number of times. So, the sum of all the numbers formed is 8 (sum of digits) * 1111 (because each digit will appear in each position) * 12 (total number of four-digit numbers) = 106656.

To solve this problem, we need to understand that each digit will appear in each position (thousands, hundreds, tens, ones) the same number of times.

1. First, let's find out how many 4-digit numbers can be formed using the digits 2, 1, 2, 3. Since the digits are not all distinct, we use the formula for permutations of a multiset: n! / (r1! * r2! * ... * rk!), where n is the total number of items, and r1, r2, ..., rk are the numbers of each type of item. Here, n = 4 (four digits), r1 = 2 (two 2s), r2 = r3 = r4 = 1 (one each of 1 and 3). So the number of 4-digit numbers is 4! / (2! * 1! * 1! * 1!) = 12.

2. Each digit will appear in each position 12 / 4 = 3 times.

3. The sum contributed by each digit when it appears in a position is the digit times the position value times the number of times it appears in that position. For example, the sum contributed by digit 2 when it appears in the hundreds position is 2 * 100 * 3 = 600.

4. The total sum contributed by all digits in all positions is:

   • Thousands position: (2 * 1000 * 3) + (1 * 1000 * 3) + (2 * 1000 * 3) + (3 * 1000 * 3) = 24000
   • Hundreds position: (2 * 100 * 3) + (1 * 100 * 3) + (2 * 100 * 3) + (3 * 100 * 3) = 2400
   • Tens position: (2 * 10 * 3) + (1 * 10 * 3) + (2 * 10 * 3) + (3 * 10 * 3) = 240
   • Ones position: (2 * 1 * 3) + (1 * 1 * 3) + (2 * 1 * 3) + (3 * 1 * 3) = 24

5. Adding up all these sums, we get 24000 + 240