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

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

A
$41 (9^{3})$
B
$37 (9^{3})$
C
$7 (9^{4})$
D
$41 (9^{4})$

Solution:

If 7 is used at first place, the number of numbers is 9⁴ and for any other four places it is 4 $\times$ 8 x 9³.

the total number of ways = $9^{3} (9 + 32) = 41 (9^{3})$

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