Answers for For how many different counting numbers between 10 and 200 is the sum of the digits equal to 6 if zero is not a digit of any numbers?

Between 10 and 100 say we have the number 10a+b where 0<a,b<10.

Now, we want numbers such that a+b = 6. Clearly there are 5 solutions: (a,b) = (1,5),(2,4),(3,3),(4,2),(5,1) corresponding to the numbers 15, 24, 33, 42 and 51.

Now between 100 and 200 we have numbers of the form 100+10a+b with 1+a+b=6 or equivalently a+b=5. Here we have 4 solutions: (a,b) = (1,4),(2,3),(3,2),(4,1) corresponding to the numbers 114, 123, 132 and 141. So all in all we have 9 solutions:

15, 24, 33, 42, 51, 114, 123, 132 and 141