Question

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

• A. $\left(41\right)\left(9^3\right)$
• B. $\left(37\right)\left(9^3\right)$
• C. $\left(7\right)\left(9^4\right)$
• D. $\left(41\right)\left(9^4\right)$

Answer 1

The correct option is A $\left(41\right)\left(9^3\right)$

Let us consider the following cases,

Case 1 : 7 is the First digit

7 __ __ __ __

Then, number of ways the remaining places could be filled $=9\cdot 9\cdot 9\cdot 9$

Case 2 : 7 is the second digit

__ 7 __ __ __

Then, number of ways the remaining places could be filled $=8\cdot 9\cdot 9\cdot 9$. (As $0$ can not be the first digit)

Case 3 : 7 is the third digit

__ __ 7 __ __

Then, number of ways the remaining places could be filled $=8\cdot 9\cdot 9\cdot 9$ (As $0$ can not be the first digit)

Case 4 : 7 is the fourth digit

__ __ __ 7 __

Then, number of ways the remaining places could be filled $=8\cdot 9\cdot 9\cdot 9$. (As $0$ can not be the first digit)

Case 5 : 7 is the last digit

__ __ __ __ 7

Then, number of ways the remaining places could be filled $=8\cdot 9\cdot 9\cdot 9$. (As $0$ can not be the first digit)

$\therefore$ The total number of such five digit numbers $=9\cdot 9\cdot 9\cdot 9+8\cdot 9\cdot 9\cdot 9+8\cdot 9\cdot 9\cdot 9+8\cdot 9\cdot 9\cdot 9+8\cdot 9\cdot 9\cdot 9$

$=9^4+4\cdot 8\cdot 9^3=(9+32)\cdot 9^3=41\cdot 9^3$

Hence, option (A) is correct.