Student is to answer 7 out of 10 questions in an examination 
Student is to answer 7 out of 10 questions in
  an examination. How
  many if she must answer at least 3 of the first 5
  questions?

Credit: A FIRST COURSE IN PROBABILITY - Sheldon Ross
University of Southern California
The answer is:
$$\binom{5}{3}\binom{5}{4}+\binom{5}{4}\binom{5}{3}+\binom{5}{5}\binom{5}{2}=110$$
Why my idea is not good?
$$\binom{5}{3}\binom{7}{4}=350$$
I take 3 of 5 of the first questions, and then I have another 7 questions that I can answer and I choose 4. Thank you!

You are counting those cases in which she answers four or five of the first five questions more than once.  You are counting them both among the $\binom{5}{3}$ selections of three of the first five questions and among the $\binom{7}{4}$ selections of four additional questions.
Suppose she answers four of the first five questions.  You have counted that case four times, once for each of the $\binom{4}{3}$ ways you could have selected three of those four questions as a member of the three questions that you selected from the first five.
Similarly, if she answers all five of the first five questions, you have counted that case $10$ times, once for each of the $\binom{5}{3}$ ways you could have selected three of those five questions as a member of the three questions that you selected from the first five. 
Note that
$$\binom{3}{3}\binom{5}{3}\binom{5}{4} + \binom{4}{3}\binom{5}{4}\binom{5}{3} + \binom{5}{3}\binom{5}{5}\binom{5}{2} = 350 = \binom{5}{3}\binom{7}{4}$$
