Question

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

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

Answer 1

Answer: (A)

(i) Case I $\to 7XXXX$
at places marked $X$, we can place 0 to 9 , except 7 , in $9\times 9\times 9\times 9=9^4$ ways
(ii) Case II $\to XY\underline{YY}$
If 7 is at some other place, first place marked $X$ can be filled in 8 ways ( 0 and 7 not allowed) and places marked $Y$ can be filled in 9 ways. 7 can be placed at any of the four places in $4\times 8\times 9^3$ ways
$\Rightarrow$ Required number of ways $=9^4+32\times 9^3=41\left(9^3\right)$