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Hint: In the question itself it is given to count the number of three-digit numbers having no digit as 5. Thus, first we will be figuring out the available options that are valid for each place of the three-digit number and then we will be using the Fundamental principle of counting.

Consider the units place of the three-digit number. Clearly given in the question that the number cannot have 5 as any of the digit, thus then digit 5 is neglected. Thus, the unit place of the three-digit number can have values 0,1,2,3,4,6,7,8,9, which are 9 in total.

Moving on to the ten’s place, due to the similar reason we can say that the ten’s place can have the digits 0,1,2,3,4,6,7,8,9 which are also 9 in total.

Lastly, we consider the hundredth place. In addition to the fact that 5 cannot come at hundredth place, also it is also important that we should note that 0 also can’t come at hundreds place. Because if it comes at the hundredth place then the number will become a two digit number which is wrong. Thus, the digits that can be at the hundreds place are 1,2,3,4,6,7,8,9, which are 8 in total.

Thus, we get that the number of possible digits at one’s, tens’ and hundredth place are 9,9, and 8 respectively.

Thus, we multiply all of them to find the number of three-digit numbers not having 5 as any digit.

$

\Rightarrow {\text{Number of possible ways}} = 9 \times 9 \times 8 \\

= 648 \\

$

Hence, there are a total of 648 three-digit numbers having no digit as 5. C is the correct option.

Note: It is highly important to consider all the possible scenarios to find the possible digits at a certain place. If you miss anything the answer would be wrong. For example, if an important point of 0 being at hundreds place will make the number a 2-digit number was missed, we might not have got the correct answer.

__Complete step-by-step solution__Consider the units place of the three-digit number. Clearly given in the question that the number cannot have 5 as any of the digit, thus then digit 5 is neglected. Thus, the unit place of the three-digit number can have values 0,1,2,3,4,6,7,8,9, which are 9 in total.

Moving on to the ten’s place, due to the similar reason we can say that the ten’s place can have the digits 0,1,2,3,4,6,7,8,9 which are also 9 in total.

Lastly, we consider the hundredth place. In addition to the fact that 5 cannot come at hundredth place, also it is also important that we should note that 0 also can’t come at hundreds place. Because if it comes at the hundredth place then the number will become a two digit number which is wrong. Thus, the digits that can be at the hundreds place are 1,2,3,4,6,7,8,9, which are 8 in total.

Thus, we get that the number of possible digits at one’s, tens’ and hundredth place are 9,9, and 8 respectively.

Thus, we multiply all of them to find the number of three-digit numbers not having 5 as any digit.

$

\Rightarrow {\text{Number of possible ways}} = 9 \times 9 \times 8 \\

= 648 \\

$

Hence, there are a total of 648 three-digit numbers having no digit as 5. C is the correct option.

Note: It is highly important to consider all the possible scenarios to find the possible digits at a certain place. If you miss anything the answer would be wrong. For example, if an important point of 0 being at hundreds place will make the number a 2-digit number was missed, we might not have got the correct answer.

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