Answer

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Hint: In three-digit numbers, we can place any number out of given numbers except zero. Why?

According to the given question, we have to form a three-digit number with given digits. So, essentially, we have three places to fill. Now the hack here is, if 0 will be placed on the first place then it’ll not be a three-digit number anymore. We can’t place it in the first place.

Hence, we have 5 choices for the first place which are 1, 2, 3, 7, 9. In the second and third place, we can place any given number. We have 6 choices for these places, which are 0, 1, 2, 3, 7, 9. So, the total numbers will be $5 \times 6 \times 6 = 180$ . Hence, option A will be the correct option.

Note: We might think why we are multiplying these numbers in order to get total choices. It's just because these choices are independent with each other. What we are placing at the first place, it has nothing to do with what we are placing at the second place. Whenever we have independent events then we multiply to get the total outcome.

According to the given question, we have to form a three-digit number with given digits. So, essentially, we have three places to fill. Now the hack here is, if 0 will be placed on the first place then it’ll not be a three-digit number anymore. We can’t place it in the first place.

Hence, we have 5 choices for the first place which are 1, 2, 3, 7, 9. In the second and third place, we can place any given number. We have 6 choices for these places, which are 0, 1, 2, 3, 7, 9. So, the total numbers will be $5 \times 6 \times 6 = 180$ . Hence, option A will be the correct option.

Note: We might think why we are multiplying these numbers in order to get total choices. It's just because these choices are independent with each other. What we are placing at the first place, it has nothing to do with what we are placing at the second place. Whenever we have independent events then we multiply to get the total outcome.

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