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
 A: 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 
$\left( \binom{5}{1}+\binom{5}{1}+\binom{5}{1}\right)\cdot \left(\binom{5}{2}\cdot \binom{5}{2}+ \binom{5}{1}\cdot \binom{5}{3}\right)=2250$
A: Alternate way to solve:
Total ways to select 5 questions without restrictions: 15C5 = 3003
Case 1: We select all the 5 questions from only one section.

*

*From Section A only: 5C5 = 1

*From Section C only: 5C5 = 1

*From Section B only: 5C5 = 1

Sub Total = 3
Case 2:
We select 4 questions from only one section.

*

*4 from Section A and 1 from Section B: (5C4)(5C1)

*And Vice-versa: 1 from Section A and 4 from Section B: (5C1)(5C4)

*4 from Section B and 1 from Section C: (5C4)(5C1)

*And Vice-versa: 1 from Section B and 4 from Section C: (5C1)(5C4)

*4 from Section C and 1 from Section A: (5C4)(5C1)

*And Vice-versa: 1 from Section C and 4 from Section A: (5C1)(5C4)

Adding Everything together: 6 * (5C4)(5C1) = 150
Case 3:
We select 3 questions from one section and 2 questions from another and not selecting anything form the third section.

*

*3 from Section A and 2 from Section B: (5C3)(5C2)

*And Vice-versa: 2 from Section A and 3 from Section B: (5C2)(5C3)

*3 from Section B and 2 from Section C: (5C3)(5C2)

*And Vice-versa: 2 from Section B and 3 from Section C: (5C2)(5C3)

*3 from Section C and 2 from Section A: (5C3)(5C2)

*And Vice-versa: 2 from Section C and 3 from Section A: (5C2)(5C3)

Adding Everything together: 6 * (5C3)(5C2) = 600
Finally, Case 1, Case 2 and Case 3 are to be rejected as they are not complying the conditions.
Required no of ways = Total - (Case 1 ways) - (Case 2 ways) - (Case 3 ways) = 3003 - 3 - 150 - 600 = 2250
