Question

Find the total number of numbers less than $3\times {{10}^{8}}$ that can be formed using the digits 1, 2, 3.
A. $\dfrac{1}{2}\left( {{3}^{9}}+4\times {{3}^{8}} \right)$
B. $\dfrac{1}{2}\left( {{3}^{9}}-3 \right)$
C. $\dfrac{1}{2}\left( 7\times {{3}^{8}}-3 \right)$
D. $\dfrac{1}{2}\left( {{3}^{9}}-3+{{3}^{8}} \right)$

Answer 1

We first find the numbers’ characteristics which are less than $3\times {{10}^{8}}$. We find the number of ways 1 to 8 digit’s number can be made which are less than $3\times {{10}^{8}}$. For the 9-digit numbers we take special care about the first digit. We add up the numbers and find the solution of the problem.

Complete step-by-step answer:
We need to find numbers less than $3\times {{10}^{8}}$. In $3\times {{10}^{8}}$, there are eight zeroes and the number itself is nine digits. So, any number having digits’ number less than nine will be eligible to be less than $3\times {{10}^{8}}$ for all cases. But for nine-digit numbers we need to take special care as the starting number can’t be 3.
So, we care about the numbers having numbers of 1 to 8.
So, let’s assume we are picking number r from 1 to 8. This means we have to fill up r places with possible options being 1, 2, 3.
There are no limitations to how many times we can use the digits 1, 2, 3.
So, the number of ways or places can be filled up with 3 possible options with repetition being allowed is ${{3}^{r}}$.
r will vary from 1 to 8 as there are 8 different possible places a number can have to be less than $3\times {{10}^{8}}$.
So, we will take the sum of ${{3}^{r}}$, r running from 1 to 8. That will be equal to
$\sum\limits_{r=1}^{8}{{{3}^{r}}}={{3}^{1}}+{{3}^{2}}+..+{{3}^{8}}=3\dfrac{{{3}^{8}}-1}{3-1}=\dfrac{3}{2}\left( {{3}^{8}}-1 \right)$.
Now we will find a nine digits number. The first digit has to be either 1 or 2 and the rest can be anything. So, the number of ways will be $2\times {{3}^{8}}$.
So, total number of ways are $\dfrac{3}{2}\left( {{3}^{8}}-1 \right)+2\times {{3}^{8}}=\dfrac{1}{2}\left( 3\times {{3}^{8}}-3+4\times {{3}^{8}} \right)=\dfrac{1}{2}\left( 7\times {{3}^{8}}-3 \right)$.

So, the correct answer is “Option C”.

Note: We also can find the number of ways 1 to 9 digit’s number can be made which are less than $3\times {{10}^{8}}$. Then we eliminate the 9-digit numbers which start with 3 as those numbers will be greater than $3\times {{10}^{8}}$. So, first step can be done in $\sum\limits_{r=1}^{9}{{{3}^{r}}}=3\dfrac{{{3}^{9}}-1}{3-1}=\dfrac{3}{2}\left( {{3}^{9}}-1 \right)$. Second step will be done in ${{3}^{8}}$ ways. The answer will be $\dfrac{3}{2}\left( {{3}^{9}}-1 \right)-{{3}^{8}}=\dfrac{1}{2}\left( 9\times {{3}^{8}}-3-2\times {{3}^{8}} \right)=\dfrac{1}{2}\left( 7\times {{3}^{8}}-3 \right)$.