Question

Total number less than $3\times {10}^8$ and can be formed using the digits $1,2,3$ is equal to

• A. $\frac{1}{2}(3^9+4\times 3^8)$
• B. $\frac{1}{2}(3^9-3)$
• C. $\frac{1}{2}(7\times 3^8-3)$
• D. $\frac{1}{2}(3^9-3+3^8)$

Answer 1

The correct option is C $\frac{1}{2}(7\times 3^8-3)$

The number formed is less than $3\times {10}^8=300,000,000$, i.e., the number can have at most 9 digits.
Now there will 3 exactly 3 number than can be formed using the digits 1, 2, or 3 and has only one digit.
To form the 2-digit numbers using the digits, 1, 2, or 3, there will be 3 choices for each of place(tens and units) of the number. Hence there will be $3^2$ such numbers.
Similarly there will be $3^3,3^4,3^5,3^6,3^7$ and $3^8$ numbers that will have $3,4,5,6,7$ and $8$ digits and formed using 1, 2 or 3.
For 9 digit numbers, we have only 2 possibilities for the left most digit as the number formed are less than $3\times {10}^8$. The remaining 8 digits can be filled in $3^8$ ways.
Hence the total numbers less than $3\times {10}^8$ and formed using the digits 1, 2 and 3 are
$3+3^2+3^3+3^4+3^5+3^6+3^7+3^8+2\times 3^8$
$=\frac{3(3^8-1)}{3-1}+2\times 3^8$
$=\frac{3^9-3+4\times 3^8}{2}$
$=\frac{7\times 3^8-3}{2}$