Question

How many natural numbers less than a million can be formed using the digits \[0,7{\text{ }}and{\text{ }}8\]?
$\left( a \right){\text{ 728}}$
$\left( b \right){\text{ 729}}$
$\left( c \right){\text{ 485}}$
$\left( b \right){\text{ 486}}$

Answer 1

Hint:
Since for the natural numbers less than a million the number should be made with these digits and should be greater than zero and less than a million. So we will have a maximum of six digits and we fill these places with \[0,7{\text{ }}and{\text{ }}8\] digits.

Complete step by step solution:
So from the question, we have the following information:
For a number to be more prominent than zero it must have at any rate a solitary non-zero digit.
For a number to be not exactly a million it can have a limit of $6$ digits.
Presently, we have $6$ places, each spot can be loaded up with $3$ digits \[0,7{\text{ }}and{\text{ }}8\].
So we will have the numbers given as
\[ \Rightarrow 3 \times 3 \times 3 \times 3 \times 3 \times 3 = {3^6}\]
But as we know that the one pattern in these above all digits being filled with zero
So the numbers will be ${3^6} - 1$.
So on solving it, we get the numbers as
$ \Rightarrow 729 - 1 = 728$
Therefore, the total numbers are$728$.

Therefore, the option $\left( a \right)$ will be correct.

Note:
There is one more way to calculate it. For this, first of all, we will find the total number of six-digit numbers which will be $2 \times 3 \times 3 \times 3 \times 3 \times 3 = 486$
Similarly, for five digits numbers, it will be $162$, and for four-digit numbers, it will be $54$ and similarly, for $3, 2, 1$ digit numbers, it will be $18, 6, 2$ respected. Now on adding all the numbers that are $486 + 162 + 54 + 18 + 6 + 2 = 728$, we will get the total numbers. So in this way also we can find the total numbers. So for solving this only concept needed is how we arrange the digits and then we can simply calculate and answer it. And by using the first method we can also reduce the error as it requires fewer calculations.