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

Solution

Maths sir · 4 steps

Do not list all 12 numbers: find what each place column adds up to, then multiply by 1111.

1. Count the numbers. The digits 2, 1, 2, 3 have the 2 repeated, so the number of arrangements is 4!/2! = 12.
2. One place column. Fix a digit in the units place and arrange the other three. With 1 there, 2, 2, 3 give 3!/2! = 3 ways. With 3 there, 1, 2, 2 also give 3 ways. With 2 there, 1, 2, 3 are all different and give 3! = 6 ways. The check 3 + 3 + 6 = 12 confirms every number is counted.

   Units sum = 1 × 3 + 3 × 3 + 2 × 6 = 24

3. All four places. The tens, hundreds and thousands columns give the same sum, 24, by the same argument. So the total is 24 × (1 + 10 + 100 + 1000) = 24 × 1111.
4. So. The sum of all the four-digit numbers is 24 × 1111 = 26664.

The common slip: 53328. Treating the two 2s as different digits gives 24 numbers and a total of 48 × 1111 = 53328. Swapping identical 2s does not make a new number, so divide by 2!.

The idea to keep. To add all arrangements of some digits, find the sum of one place column, then multiply by 111…1. A repeated digit appears in each place more often than the other digits do.

Answer from the NTA final answer key. Solution by PrepWiser.