Question

Number of five-digit numbers that contain digit $7$ exactly once is

• A. $41\times 9^4$
• B. $33\times 9^4$
• C. $41\times 9^3$
• D. $33\times 9^3$

Answer 1

The correct option is C $41\times 9^3$

Case $1:$ When $7$ is at first place i.e., at leftmost.
Then there are $9$ choices for each of the remaining places.
Hence, the possible number that can be formed in this case is $9^4.$

Case $2:$ When $7$ is not at first place
For any other four places, it is $8\times 9^3$ and also $7$ can be arranged in $4$ places.
Hence, the possible number that can be formed in this case is $4\times 8\times 9^3.$

Hence, total possible ways $=9\times 9^3+4\times 8\times 9^3=41\times 9^3$