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The following is a simple combinatorics question:

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

The correct answer is: $^7C_3\cdot^5C_5+ ^7C_4\cdot^5C_4+^7C_5\cdot^5C_3=210+175+35=420$

But I thought of the following approach:

Since at least 3 of each type are to be attempted, so why not first choose 3 of each type first, this when counted gives: $^7C_3\cdot^5C_3=350$. Now out of 12 only 6 are left to be chosen out of, where 2 are to be chosen, so this when counted gives $^6C_2=15$. Thus by addition principle it follows that total total such cases should be $375$ which is clearly wrong. So I ask why is it wrong? Where is the fault in logic? It's obvious I'm not counting some case so what might they be?

Shyam Tripathi
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    Write your argument out. IE Suppose the questions are 1, 2, 3, 4, 5, 6, 7, A, B, C, D, E. And we chose 1, 2, 3, A, B, C. What happens next? According to you, why would we use the additional principle? – Calvin Lin Dec 02 '23 at 15:45
  • It would be the product rule of counting not the addition principle, although the product rule wouldn't work because you'd be counting things more than once. – Adam Rubinson Dec 02 '23 at 16:10
  • As has been already indicated by the comments, the problem is over-counting. The traditional remedy, which would salvage your approach, is Inclusion-Exclusion. See this article for an introduction to Inclusion-Exclusion. Then, see this answer for an explanation of and justification for the Inclusion-Exclusion formula. ...See next comment – user2661923 Dec 02 '23 at 16:12
  • You have two errors: (a) adding $^6C_2=15$ when you should be multiplying (if this method worked) and (b) a method which overcounts cases. (b) does not become so obvious until you correct (a) – Henry Dec 02 '23 at 16:13
  • However, for this particular problem, Inclusion-Exclusion, while valid, is something of a nightmare. There are $~\displaystyle \binom{7}{3} \times \binom{5}{3} = 350~$ different ways of selecting which $~6~$ questions must be included. So, you would be attempting to enumerate $$| ~S_1 \cup S_2 \cup \cdots \cup S_{350} ~|.$$ To make matters worse, the various types of intersections, and their enumerations, are variable. So, for example, you will not have that $$|S_1 \cap S_2| = |S_1 \cap S_3|.$$ So, this approach is a nightmare. – user2661923 Dec 02 '23 at 16:17
  • Suppose that from one section you ended up choosing questions $a$ through $d$. Would it be relevant to this counting problem which three of them had been among your initial $3+3$ choices and which one had been among the final pair you chose to round out your eight? – Paul Tanenbaum Dec 02 '23 at 16:51
  • Any time you find yourself saying something like "first choose 3 of each type" in a combinatorics exercise, where you say "first" because later you're going to choose more items from the same pool of items later, it's a huge red flag that you're likely making a mistake. – David K Dec 02 '23 at 18:05

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