# How many ways he can attempt the paper?

Joey has to attempt a question paper that has 3 sections with 6 questions in each section .

If he has to attempt any 8 questions , choosing atleast 2 questions from each section.

Then in how many ways he can attempt the paper?

I tried by taking 2 question in each section as

6C2 ways.

So in all 3 section , he can select 6 questions in 6C2 * 6C2 * 6C2 ways.

Now remaining 2 question he can select from remaining 12 question in 12C2 ways.

So total ways are 6C2 * 6C2 * 6C2 * 12C2 ways.

But the answer is not coming correct.

Is my approach incorrect?

Or i am missing any cases?

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You have counted many cases several times. E.g., if in the end from section 1 questions $1.1$, $1.2$, and $1.3$ were selected this has been counted three times, namely as $(1.1, 1.2)$ in the first round, $1.3$ in the second round, then as $(1.2,1.3)$ in the first round and $(1.1)$ in the second round, and finally as $\ldots\$. –  Christian Blatter Oct 4 '12 at 10:22
@can you tell me another way of solving it. –  vikiiii Oct 4 '12 at 10:30

First, let's split the problem to subproblems depending on how many questions do we choose from each section.

For example "2 questions from the first section, 4 questions from the second section, 2 questions from the third section" would be one such subproblem. How many are they?

With 8 questions, 3 sections, and at least 2 questions per section, our choices are these: 2+2+4; 2+3+3; 2+4+2; 3+2+3; 3+3+2; 4+2+2. There are essentially just two subproblems, 2+2+4 and 2+3+3, each of them repeated 3 times.

Now, in how many ways can Joey attempt each of these subproblems?

To choose 2+2+4 questions he has 6C2 * 6C2 * 6C4 options. To choose 2+3+3 questions he has 6C2 * 6C3 * 6C3 options. The answer is: 3 * 6C2 * 6C2 * 6C4 + 3 * 6C2 * 6C3 * 6C3.

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There are $\binom{18}8$ ways to choose $8$ of the $18$ questions. Let’s count the number that are not acceptable.
There are $\binom{12}8$ ways to choose them so as to have none from the first section, and another $\binom61\binom{12}7$ ways to choose them so as to have exactly $1$ from the first section. Thus, there are $\binom{12}8+\binom61\binom{12}7$ ways to choose them so as to have fewer than $2$ from the first section. There are just as many ways to choose fewer than $2$ from the second section, and just as many again to choose fewer than $2$ from the third section, so as a first approximation there are $$3\left(\binom{12}8+\binom61\binom{12}7\right)\tag{1}$$ unacceptable choices.
However, $(1)$ is too big, because we counted some unacceptable choices twice. Specifically, any set of $8$ questions that includes fewer than $2$ from two sections got counted as bad for each of those sections, so it got counted twice in $(1)$. To correct for this, we must subtract from $(1)$ the number of unacceptable sets that have too few questions from two sections. Such a set must have all $6$ questions from one section and one question from each of the other two sections. There are $3$ ways to choose the section that contributes $6$ questions and then $6^2$ ways to choose one question from each of the other two sections, so there are $3\cdot6^2$ sets that have fewer than $2$ questions from each of two sections. Thus, $(1)$ should be corrected to $$3\left(\binom{12}8+\binom61\binom{12}7\right)-3\cdot6^2\;.\tag{2}$$ Since it’s not possible to choose $8$ questions and get fewer than $2$ from each of the three sections, $(2)$ is the number of unacceptable choices. The desired result is therefore $$\binom{18}8-3\left(\binom{12}8+\binom61\binom{12}7\right)-3\cdot6^2\;;$$ if I’ve done the arithmetic correctly, this is $43~758$.