The sum of all the 4-digit distinct numbers that can be formed with the digits 1, 2, 2 and 3 is :

Solution

Maths sir · 4 steps

Do not list all 12 numbers. Count how often each digit sits in one place, then multiply by 1111.

1. How many numbers. Four digits with 2 repeated give 4!/2! = 12 distinct numbers.
2. Count each digit. Fix a digit in the units place and arrange the other three. With 1 fixed, 2, 2, 3 give 3!/2! = 3 numbers. With 3 fixed, 1, 2, 2 also give 3. With 2 fixed, 1, 2, 3 give 3! = 6. That makes 3 + 3 + 6 = 12, as expected.
3. Sum of one place. The units digits of all 12 numbers add to 3 × 1 + 6 × 2 + 3 × 3 = 24. The tens, hundreds and thousands places work the same way, so each place also adds to 24.

   S = 24 × (1000 + 100 + 10 + 1) = 24 × 1111 = 26664

4. So. The sum of all the distinct numbers is 26664, which is option (a).

The idea to keep. For the sum of all numbers, find the sum of the digits in one place and multiply by 111…1. With repeated digits, count each digit's appearances separately and do not treat the copies as different.

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