Question

Suppose n is an integer such that the sum of digits of n is 2; and 10${}^{10}$<n<10${}^{11}$. The number of different values of n is?

• A. 11
• B. 10
• C. 9
• D. 55

Answer 1

The correct option is A

11

It is given that the number has 11 digits which add up to 2

When the first digit is 1,

The number can be formed from 9 zeroes and 2 ones which can be arranged in $\frac{10!}{9!}\times$1! = 10 ways

When the first digit is 2, we can have 1 more number. Total cases = 11. Answer is option (a)