question and your explanation both are amazing!! Thank You @Deepak Poonia sir

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A student is given an exam consisting of $8$ essay questions divided into $4$ groups of $2$ questions each. The student is required to select a set of $6$ questions to answer, including at least $1$ question from each of the $4$ groups. How many sets of questions satisfy this requirement?

8 votes

$\text{Method 1:}$

$$\left(\begin{array}{l}8 \\6\end{array}\right)=28$$

Then let's take away the choices where we don't cover all $4$ groups. There are only $4.$ Why? To not cover all $4$ groups but to choose $6$ means to pick all questions but $2$ questions of the same group. That is, from $4$ groups, pick $1$ group to exclude or equivalently pick $3$ groups to include

$$\begin{gathered} \left(\begin{array}{l} 4 \\3 \end{array}\right)=\left(\begin{array}{l} 4 \\ 1 \end{array}\right)=4 \\ \therefore\left(\begin{array}{l} 8 \\ 6 \end{array}\right)-\left(\begin{array}{l} 4 \\ 3 \end{array}\right)=\left(\begin{array}{l} 8 \\ 6

\end{array}\right)-\left(\begin{array}{l} 4 \\ 1 \end{array}\right)=28-4=24 \end{gathered} $$

$\text{Method 2:}$

Because we only have four groups and you need to pick up 6 balls which means you need to pick up two groups of balls and one ball each from the other two groups so the final result would be $\left(^{4} \mathrm{C} _{2}\right) \times 2 \times 2$ which is $24.$

$\text{Method 3:}$

The student can choose two questions to omit, not both in the same group. There are $8$ options for the first question, and then there are $6$ options left for the second question. Of course the order in which the questions are chosen doesn't matter, so we get $8 \times 6 / 2=24$ options.

Alternatively, the student can choose two groups to omit a question from, and then one question from each group. This way we get $\left(^{4} \mathrm{C}_{2}\right)\left(^{2} \mathrm{C} _{1}\right)\left(^{2} \mathrm{C} _{1}\right)=6 \times 2 \times 2=24$ options.

$$\left(\begin{array}{l}8 \\6\end{array}\right)=28$$

Then let's take away the choices where we don't cover all $4$ groups. There are only $4.$ Why? To not cover all $4$ groups but to choose $6$ means to pick all questions but $2$ questions of the same group. That is, from $4$ groups, pick $1$ group to exclude or equivalently pick $3$ groups to include

$$\begin{gathered} \left(\begin{array}{l} 4 \\3 \end{array}\right)=\left(\begin{array}{l} 4 \\ 1 \end{array}\right)=4 \\ \therefore\left(\begin{array}{l} 8 \\ 6 \end{array}\right)-\left(\begin{array}{l} 4 \\ 3 \end{array}\right)=\left(\begin{array}{l} 8 \\ 6

\end{array}\right)-\left(\begin{array}{l} 4 \\ 1 \end{array}\right)=28-4=24 \end{gathered} $$

$\text{Method 2:}$

Because we only have four groups and you need to pick up 6 balls which means you need to pick up two groups of balls and one ball each from the other two groups so the final result would be $\left(^{4} \mathrm{C} _{2}\right) \times 2 \times 2$ which is $24.$

$\text{Method 3:}$

The student can choose two questions to omit, not both in the same group. There are $8$ options for the first question, and then there are $6$ options left for the second question. Of course the order in which the questions are chosen doesn't matter, so we get $8 \times 6 / 2=24$ options.

Alternatively, the student can choose two groups to omit a question from, and then one question from each group. This way we get $\left(^{4} \mathrm{C}_{2}\right)\left(^{2} \mathrm{C} _{1}\right)\left(^{2} \mathrm{C} _{1}\right)=6 \times 2 \times 2=24$ options.

how to count the number of cases that are overcounted? I have done it for one case, is it correct?

@Vaibhav04 Your analysis is Absolutely Correct and Beautiful.

You have understood this concept as clearly as anyone can. Keep it up.

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