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Q.
The sum of all the numbers of four different digits that can be made by using the digits $0, 1, 2$, and $3$ is
Permutations and Combinations
Solution:
The number of numbers with $0$ in the unit’s place is $3! = 6$.
The number of numbers with $1$ or $2$ or $3$ in the unit’s place is $3! - 2! = 4$. Therefore, the sum of the digits in the unit’s place is $6 \times 0 + 4 \times 1 + 4 \times 2 + 4 \times 3 = 24$.
Similarly, for the ten’s and hundred’s places, the number of numbers with $1$ or $2$ in the thousand’s place is $3!$ Therefore, the sum of the digits in the thousand’s place is $6 \times 1 + 6 \times 2 + 6 \times 3 = 36$.
Hence, the required sum is $36 \times 1000 + 24 \times 100 + 24 \times 10 + 24$.