Question

Number of different natural numbers which are smaller than two hundred million and using the digit 1 or 2 is:
A. ${3.2^8} - 2$
B. $(3){.2^8} - 1$
C. $(3){.2^9} - 1$
D. None

Answer 1

Hint: According to the question we have to determine the number of different natural numbers which are smaller than two hundred million and using the digit 1 or 2. So, first of all as we know that the possible numbers which are smaller than two hundred million and using the digit 1 or 2 can be either 1 digit, 2 digits, 3 digits, 4 digits, ……………., 9 digits.
Now, we have to determine the numbers of such 9 digits numbers and then the numbers of such 8 digits numbers same as the numbers of such 7 digits numbers and the numbers of such 6 digits numbers and so on till the numbers of such 1 digit number.
Now, we have to add all the numbers obtained thus the numbers that are smaller than 200000000 and we have to use only 1 or 2.
Now, to find the sum of all the numbers we have to use the formula to find the sum as mentioned below:

Formula used: $ \Rightarrow \sum\limits_{n = 1}^{n = 8} {{x^n}} .................(A)$
Now, to solve the obtained expression we have to use the formula which is mentioned below:
$ \Rightarrow \sum\limits_{k = 1}^n {{r^{k - 1}}} = \dfrac{{1 - {r^n}}}{{1 - r}}.........................(B)$
Now, on substituting all the values in the formula (B) above we can easily determine the number of different natural numbers which are smaller than two hundred million and using the digit 1 or 2.

Complete step by step solution:
Step 1: First of all as we know that the possible numbers which are smaller than two hundred million and using the digit 1 or 2 can be either 1 digit, 2 digits, 3 digits, 4 digits, ……………., 9 digits.
Step 2: Now, as from the step 1 we have to determine the all of those numbers which are as below:
The numbers of such 9 digits numbers are ${2^8}$,
The numbers of such 8 digits numbers are ${2^8}$,
The numbers of such 7 digits numbers are ${2^7}$,
The numbers of such 6 digits numbers are ${2^6}$,
The numbers of such 5 digits numbered are ${2^5}$, and so on.
Step 3: Now, we have to add all the numbers as obtained in the solution step 2. Hence,
$ \Rightarrow {2^8} + {2^8} + {2^7} + {2^6} + {2^5} + {2^4} + {2^3} + {2^2} + {2^1}$………………….(1)
Sep 4: Now, to add all the numbers as obtained (1) in the solution step 3 we have to use the formula (A) as mentioned in the solution hint.
$ \Rightarrow {2^8} + {2^8} + {2^7} + {2^6} + {2^5} + {2^4} + {2^3} + {2^2} + {2^1} = {2^8} + \sum\limits_{k = 1}^8 {{2^k}} $…………………(2)
Step 5: Now, we have to rearrange the terms as obtained (2) in the solution step 4. Hence,
$ \Rightarrow {2^8} + \sum\limits_{k = 1}^8 {{2^k}} = {2^8} + 2\sum\limits_{k = 1}^8 {{2^{k - 1}}} $……………………..(3)
Step 6: Now, to solve the expression (3) as obtained in the solution step 3 we have to use the formula (B) as mentioned in the solution hint.
$ \Rightarrow {2^8} + 2\sum\limits_{k = 1}^8 {{2^{k - 1}}} = {2^8} + 2\dfrac{{1 - {2^8}}}{{1 - 2}}$
On solving the expression as obtained just above,
$
   = {2^8} + 2({2^8} - 1) \\
   = 3.({2^8} - 2)
 $
Final solution: Hence, with the help of the formula (A) and (B) we have determined the number of different natural numbers which are smaller than two hundred million and using the digit 1 or 2 is $ = 3.({2^8} - 2)$.

Therefore option (A) is correct.

Note: The possible numbers which are smaller than two hundred million and using the digit 1 or 2 can be either 1 digit, 2 digits, 3 digits, 4 digits, ……………., 9 digits.
The numbers of such 9 digits numbers are ${2^8}$, the numbers of such 8 digits numbers are ${2^8}$, the numbers of such 7 digits numbers are ${2^7}$, The numbers of such 6 digits numbers are ${2^6}$, ………………., and The numbers of such 1 digit number is ${2^1}$.