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# How many 3 – digit numbers can be formed using the digits 0, 1, 3, 5, 7 while each digit may be repeated any number of times?

Last updated date: 09th Aug 2024
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Hint: As we have to form 3 – digit numbers, the first digit cannot be 0. So, the first digit can be arranged in 4 ways. Now, as repetition is allowed, the second and the third digit can be arranged in 5 ways. The total numbers formed will be equal to the product of these values.

Complete step-by-step solution:
In this question, we are asked how many 3 digit numbers can be formed using the digits 0, 1, 3, 5, 7 where repetition is allowed.
Given digits: 0, 1, 3, 5, 7
Therefore, there are a total of 5 digits.
Now, as we have to form 3 digit numbers, let us draw three boxes.

First box will be hundreds, the second box will be tens and the third box will be ones.
Now, as the number is 3 – digit, the first box cannot use digit 0, because if we use 0 as first digit, the number will not be a 3 digit number. So, only 4 digits that are 1, 3, 5 and 7 can be used in the first box.

Therefore, there are 4 ways to fill the first box.
Now, for the second box, any out of the 5 given digits can be used to fill the box as repetition is allowed.
Therefore,

And for the third box too, we can fill it using any of the 5 given digits. Therefore, we get

So, therefore the total ways for forming 3 digit number will be
$\Rightarrow$Total ways of arrangement $= 4 \times 5 \times 5 = 100$
Therefore, a total of $100$ 3 digit numbers can be formed using the digits 0, 1, 3, 5, 7 when repetition is allowed.

Note: Here, note that if repetition were not allowed, then we would have got different answers.
For first digit:
As we have to form a 3 digit number, we cannot use 0 as the first digit. So,

For second digit:
Now, suppose we used 1 as the first digit, so now it cannot be used again for the second and third digit. So, the second digit can be 0, 3, 5 or 7. Therefore,

For third digit:
Suppose we used 3 as the second digit, so now 3 cannot be used for the third digit. So, the third digit can be 0, 5 or 7. Therefore,

$\Rightarrow$Total numbers formed when repetition is not allowed$= 4 \times 4 \times 3 = 48$