Where is my error in this combinations word problem? 
A student has to answer $10$ out of $13$ questions in an examination. The number of ways in which he can answer if he must answer at least $3$ of the first five questions is:

The answer is not $^5C_3 \times {^{10}C_7}$. Why?????
$C$ here is combination.
 A: Welcome to MSE!
Given the first five questions:
$ 1_q \:\: 2_q \:\: 3_q \:\: 4_q \:\: 5_q $
these are all the $10$ possibilities of choosing $3$ out of the $5$:
$\{1_q,2_q,3_q\}, \{1_q,2_q,4_q\}, \{1_q,2_q,5_q\} , \{1_q,3_q,4_q\} , 
\{1_q, 3_q, 5_q \}, \{ 1_q, 4_q, 5_q \} $
$\{2_q, 3_q, 4_q\} , \{2_q, 3_q, 5_q \} , \{2_q , 4_q, 5_q\}$
$\{3_q,4_q,5_q\}$
say the student answered the first $3$ out of the $5$ : $\{1_q,2_q,3_q\}$
‘then there are still’ $10$ questions left to choose from $\{4_q,5_q,...,13_q\}$ and the student answered the $7$ questions $4_q, 6_q, 7_q, 8_q, 9_q,10_q, 11_q$,
now say the student answered questions $1_q,3_q,4_q$ out of the $5$ first,
and then answered the $7$ more questions $2_q, 6_q, 7_q, 8_q, 9_q,10_q,11_q$,
in essence the student answered the same set of questions $\{1_q,2_q,3_q,4_q,6_q,7_q,8_q,9_q,10_q,11_q\}=\{1_q,3_q,4_q,2_q,6_q,7_q,8_q,9_q,10_q,11_q\}$
but just in a different order, but that is double counting nonetheless.
The key here is at least $3$ of the first $5$ questions, the student must answer $3$, any $3$, but they may answer more, and you could want to break it up into exactitudes, the student answers exactly $3$ of the first $5$, exactly $4$ of the first $5$, all $5$ of the first $5$ to avoid overlapping/double-counting.
Say the student answers exactly $3$ of the first five questions:
$$\binom{5}{3}\binom{8}{7}$$
The student may also answer exactly $4$ of the first five questions:
$$\binom{5}{4}\binom{8}{6}$$
and the student may also answer all $5$ of the first five questions:
$$\binom{5}{5}\binom{8}{5}$$
all possibilities is the sum of all these $\binom{5}{3}\binom{8}{7} +\binom{5}{4}\binom{8}{6}+\binom{5}{5}\binom{8}{5}=276$.
A: Your solution implies the exam taker answers $3$ questions from the first $5$, and then $7$ questions from the remaining $10$.
This counts some of the combinations more than once.
For example, label the first $5$ question $A,B,C,D,E$, and the last $8$ as $a,b,c,d,e,f,g,h$.
Then you count say $ABCDEabcde$ several times, e.g.:
(ABC)DEabcde, (ABD)CEabcde, (CDE)ABabcde, etc...
In fact it counts each one $\binom{5}{3}=10$ times.
A different solution is to notice that if you answer $10$ questions from $13$, then, by the Pigeonhole Principle, you must have answered at least $2$ from the first $5$, and there are $\binom{5}{2}=10$ ways to do that. So the answer is $\binom{13}{10}-\binom{5}{2}=286-10=276$.
A: This is similar to the Textbook question in India if you recollect. First, you divide the number of questions into two categories- the first $5$ questions and the remaining $8$ questions.
Now pay attention to the words of the question. it says

" he must answer atleast 3 questions..."

and of course

student has to answer 10 questions

So the student may any $n$ number of questions among the first 5 and any $m$ number of questions among the last 8. The only conditions given are

n is greater than 3 and at most 5


$m+n=10$

You can work out the other number of possibilities by yourself and add them up
(this was supposed to be a comment)
