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Question
Suppose n is an integer such that the sum of digits of n is 2; and 1010<n<1011. The number of different values of n is?
Options:
A .  11
B .  10
C .  9
D .  55
E .  118
Answer: Option A
:
A
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 10!9!×1! = 10 ways
When the first digit is 2, we can have 1 more number. Total cases = 11. Answer is option (a)

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