<s>[INST] There are 3 sections in a question paper with 5 questions each. There are 3 sections in a question paper each containing 5 questions. A candidate has to solve only 5 questions, choosing at least one question from each section. In how many ways can he make his choice?
I have thought of a solution but I am over counting the number of ways.
No. of ways to choose one question from each section = (5C1)^3
No. of questions remaining = 12
No. of ways to pick 2 questions from remaining 12 questions = 12 * 11
Total number of ways = (5C1)^3 * 12 * 11
Can somebody tell me where I'm going wrong
 [/INST] He can choose the following combinations of questions from each section:
$(2,2,1);(2,1,2);(1,2,2);(3,1,1);(1,3,1);(1,1,3)$
Each of the first three combinations has $\binom{5}{2}\cdot \binom{5}{2}\cdot \binom{5}{1}$ ways. And each of the next three combinations has $\binom{5}{3}\cdot \binom{5}{1}\cdot \binom{5}{1}$ ways.
In total he can make $3\cdot \left(\binom{5}{2}\cdot \binom{5}{2}\cdot \binom{5}{1}+ \binom{5}{3}\cdot \binom{5}{1}\cdot \binom{5}{1}\right)=2250$ choices.

For every section we have one path. We choose 3 question from each section. For every path we have to add ($\text{not multiply}$) the ways
$\left( \binom{5}{1}+\binom{5}{1}+\binom{5}{1}\right)$
Now we make a case decision. 
a) We choose 2 questions from one (other) section and 2 questions from the remaining section.
$ \binom{5}{2}\cdot \binom{5}{2}$
b) We choose 1 questions from one (other) section and 3 questions from the remaining section.
$ \binom{5}{1}\cdot \binom{5}{3}$
Therefore in total we have 
$\left( \binom{5}{1}+\binom{5}{1}+\binom{5}{1}\right)\cdot \left(\binom{5}{2}\cdot \binom{5}{2}+ \binom{5}{1}\cdot \binom{5}{3}\right)=2250$
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