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 $\rightarrow 7 X X X X$
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 $\rightarrow X Y \underline{Y Y}$
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 $=9^{4}+32 \times 9^{3}=41\left(9^{3}\right)$