Question

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

• A. 11
• B. 10
• C. 9
• D. 8

Answer 1

The correct option is A 11

Option (a)

There are 11 digits in this number

When sum of digits is 2, there are 2 options

There are 2 ones, with one 1 fixed in the first position

The question in now based on the arrangement of 10 zeroes and 1 one

Number of possibilities = $\frac{10!}{9!1!}$ = 10 ways

When 2 is the first digit = one possibility.

Total number of possibilities = 11