Question

Number of different natural numbers which are smaller than two hundred million and using only the digits $1$ or $2$ is

• A. $3(2^8-1)$
• B. ${.2}^8-1$
• C. $2.(2^9-1)$
• D. None

Answer 1

The correct option is A $3(2^8-1)$

These numbers will be either 1 digit,2 digits,.......9 digits.

The number of such 99 digits numbers are $2^8$,

the number of such 88 digits numbers are $2^8$ and so on.

So the total numbers are

$2^8+2^8+2^7+2^6+.....+2^2+2^1$

As we know:

$2^8+2^8+2^7+2^6+2^5+2^4+2^3+2^2+2^1=2^8+{\sum }_{k=1}^8{2}^k$

$=2^8+2{\sum }_{k=1}^8{2}^{k-1}$

Use Geometric series formula: ${\sum }_{k=1}^n{r}^{k-1}=\frac{1-r^n}{1-r}$

$=2^8+2.\frac{1-2^8}{1-2}$

$=2^8+2(2^8-1)$

$=2^8+2.2^8-2$

$=3(2^8-1)$

Option B is Correct