Techniques of counting Suppose you need to answer 10 out of 13 questions at an examinatioin. How many choices do you have if you must answer the first two questions
? 
I think out of 13 , 2 questions must answer, so remaining is 11. 
 A: Hint If you must answer the first two questions, how many questions would need to be answered out of how many remaining questions?
A: $ 11 \choose 8 $+ $ 11 \choose 9 $+ $ 11 \choose 10$+ $11 \choose 11$ as while you need to answer 10, there isn't anything stated about answering 11, 12, or all 13 questions and thus you need to cover the cases where one is answering 8 or more of the remaining questions.
To compute those values, I can take the other combinations and compute as follows:
$ 11\choose3 $+$ 11\choose2$+$11\choose1$+$11\choose0$ = $\frac{11*10*9}{2*3}+\frac{11*10}{2}+11+1= 165+55+12=232$
Thus, there are 232 possible choices if you are answering at least 10 questions on the exam.
If you are answering exactly 10 questions, then the answer 165 as the other cases can be discarded.
To work this out a bit more to show why 11 is simply too low, consider which 3 questions are being skipped in the simple 10 questions being answered case.  Let's take a few examples if we just write out which questions are being skipped:
{3,4,5}, {3,4,6}, {3,4,7}, {3,4,8}, {3,4,9}, {3,4,10}, {3,4,11}, {3,4,12}, {3,4,13},
{3,5,6}, {3,5,7}, {3,5,8}, {3,5,9}, {3,5,10}, {3,5,11}, {3,5,12}, {3,5,13}
Would be 17 possible combinations where I'm assuming the third question is skipped and at least one of the next 2 is skipped.  This can be computed by taking a few combinations and putting them together:


*

*If 3,4,5 are all skipped is one case to put aside.

*Consider that 1 of 4,5 will be skipped and one of the last 8 questions which is 16 possibilities.


Total: 17 possibilities there and that is just scratching the surface here to my mind.
What may help is to re-frame the question.  Instead of having 13 questions with 10 to answer, the first 2 questions are answered and that means there are now 11 questions left where 8 are to be answered.  How many ways could you choose 8 elements from a set of 11?
