# Different way of solving this combinatorics 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

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.