Question

Number of five digit numbers that contain digit $7$ exactly once (repetition of digits is allowed) is

• A. $41\times 9^3$
• B. $37\times 9^3$
• C. $7\times 9^4$
• D. $41\times 9^4$

Answer 1

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

If $7$ is used at first place, then the number of numbers is $9^4$ and for any other four places it is $8\times 9^3$
Hence total possible ways $=9\times 9^3+4\times 8\times 9^3=41\times 9^3$