NIMCET 2018 Question Paper with solutions

According to question, following cases arrises nine digit number (less than 2 × ${10}^8$) having only digits 1 and 2 = $2^8$
Eight digit number (less than 2 × ${10}^8$) having only digits 1 and 2 = $2^8$
Similarly, Seven digit numbers = $2^7$
Six digit numbers = $2^6$ ,
Five digit numbers = $2^5$ ,
Four digit numbers = $2^4$ ,
Three digit numbers = $2^3$ ,
Two digit numbers = $2^2$ and
One digit numbers = 2
So, number of all such natural numbers
= 2 + $2^2+2^3+2^4+2^5+26+2^7+2^8$
= 2 + 4 + 8 + 16 + 32 + 64 + 128 + 256 + 256 = 766