Different way of solving this combinatorics question.

Question

In an examination, a question paper consists of 12 questions divided into two parts i.e, Part I and Part II, containing 5 and 7 questions respectively. A student is required to attempt 8 question in all, selecting at least 3 from each part. In how many ways can a student select the questions?

I know the answer is -

(ways of selecting 3 out of 5 from I and 5 out of 7 from II) +
(ways of selecting 4 out of 5 from I and 4 out of 7 from II) +
(ways of selecting 5 out of 5 from I and 3 out of 7 from II)

$\binom{5}{3}$ $\times$ $\binom{7}{5}$ + $\binom{5}{4}$ $\times$ $\binom{7}{4}$ + $\binom{5}{5}$ $\times$ $\binom{7}{3}$ = 420

But what is the mistake in doing it in this way

(ways of selecting the minimum 3 from each part I and II
 and 2 out of the remaining 6 from both parts)

$\binom{5}{3}$$\times$$\binom{7}{3}$$\times$$\binom{6}{2}$ = 5250

Answer 1

The mistake is that it's an over counting.

The second method would count selecting questions $1,2,3$ from part 1, $1,2,3$ from part 2, and then $4$ and $5$ from part 1 again, multiple times. Look if we rearrange how we choose:

$1,4,5$ from part 1, $1,2,3$ from part 2, and $2,3$ from part 1 also.

The second method would consider these two as two different combinations while clearly they are the same.

Answer 2

The mistake in the second way is that you are essentially putting an order on the way you choose questions, which leads to double counting. For example, let's say you choose questions 1, 2, and 3 from Part I, questions 6, 7, and 8 from Part II, and then questions 4, and 5. This would be the same question set as if you chose 3, 4, 5, and 6, 7, 8, and then 1 and 2, but it's effectively counted differently in your expression.