<s>[INST] 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!
 [/INST] 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
$$\binom{3}{3}\binom{5}{3}\binom{5}{4} + \binom{4}{3}\binom{5}{4}\binom{5}{3} + \binom{5}{3}\binom{5}{5}\binom{5}{2} = 350 = \binom{5}{3}\binom{7}{4}$$
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