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In an exam, there are $12$ questions in total. He has to attempt $8$ questions in all. There are two parts: Part A and Part B of the question paper containing $5$ and $7$ questions respectively. How many ways are there to attempt the exam, such that you attempt eight questions and from each part you attempt at least three questions?
Please don't answer with the three case method. What I want is the error in my method?
My method: For three questions from each part, we do: $${5 \choose 3}*{7\choose3} $$. But 2 questions are lett, so for that, we multiply the expression by $${6\choose2} $$. Because 6 questions are left to choose from. But this does not give the answer, why?

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As you are aware, the actual number of ways to select eight questions from the examination given the restrictions that at least three questions must be selected from each part is $$\binom{5}{3}\binom{7}{5} + \binom{5}{4}\binom{7}{4} + \binom{5}{5}\binom{7}{3}$$ The alternative method you proposed of selecting three questions from part A, three questions from part B, and two additional questions counts the same selection more than once.

If we choose three questions from part A and five questions from part B, the second method counts the same selection of questions ten times since there are $\binom{5}{3}$ ways to choose three of the five selected questions from part B.

To make this concrete, suppose the selected questions are $A_1, A_2, A_3, B_1, B_2, B_3, B_4, B_5$. We can make this particular selection once. However, the second method counts this particular selection ten times. \begin{align*} & \color{red}{A_1}, \color{red}{A_2}, \color{red}{A_3}, \color{blue}{B_1}, \color{blue}{B_2}, \color{blue}{B_3}, B_4, B_5\\ & \color{red}{A_1}, \color{red}{A_2}, \color{red}{A_3}, \color{blue}{B_1}, \color{blue}{B_2}, B_3, \color{blue}{B_4}, B_5\\ & \color{red}{A_1}, \color{red}{A_2}, \color{red}{A_3}, \color{blue}{B_1}, \color{blue}{B_2}, B_3, B_4, \color{blue}{B_5}\\ & \color{red}{A_1}, \color{red}{A_2}, \color{red}{A_3}, \color{blue}{B_1}, B_2, \color{blue}{B_3}, \color{blue}{B_4}, B_5\\ & \color{red}{A_1}, \color{red}{A_2}, \color{red}{A_3}, \color{blue}{B_1}, B_2, \color{blue}{B_3}, B_4, \color{blue}{B_5}\\ & \color{red}{A_1}, \color{red}{A_2}, \color{red}{A_3}, \color{blue}{B_1}, B_2, B_3, \color{blue}{B_4}, \color{blue}{B_5}\\ & \color{red}{A_1}, \color{red}{A_2}, \color{red}{A_3}, B_1, \color{blue}{B_2}, \color{blue}{B_3}, \color{blue}{B_4}, B_5\\ & \color{red}{A_1}, \color{red}{A_2}, \color{red}{A_3}, B_1, \color{blue}{B_2}, \color{blue}{B_3}, B_4, \color{blue}{B_5}\\ & \color{red}{A_1}, \color{red}{A_2}, \color{red}{A_3}, B_1, \color{blue}{B_2}, B_3, \color{blue}{B_4}, \color{blue}{B_5}\\ & \color{red}{A_1}, \color{red}{A_2}, \color{red}{A_3}, B_1, B_2, \color{blue}{B_3}, \color{blue}{B_4}, \color{blue}{B_5}\\ \end{align*}

By similar reasoning, if we choose five questions from part A and three questions from part B, the second method counts the same selection of questions ten times since there are $\binom{5}{3}$ ways to choose three of the five selected questions from part A.

If we choose four questions from part A and four questions from part B, the second method counts the same selection sixteen times since we count the same set of four questions from part A four times for each of the $\binom{4}{3}$ ways we choose three of those four selected questions and the same set of four questions from part B four times for each of the $\binom{4}{3}$ ways we choose three of those four selected questions.

Note that $$10\binom{5}{3}\binom{7}{5} + 16\binom{5}{4}\binom{7}{4} + 10\binom{5}{5}\binom{7}{3} = \binom{5}{3}\binom{7}{3}\binom{6}{2}$$

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  • $\begingroup$ I've edited my answer to make what I'm saying more concrete. $\endgroup$ – N. F. Taussig Oct 8 '15 at 11:43
  • $\begingroup$ Okay, we see that this is happening. But why is this happening? Can we add something so that my method works? $\endgroup$ – Aditya Agarwal Oct 8 '15 at 11:45
  • $\begingroup$ As you can see, the method you attempted to use does not work since it counts the same set of selected questions every time you choose $3$ of the selected questions. To avoid overcounting, we use cases. I do not see a way to make the method you attempted work. $\endgroup$ – N. F. Taussig Oct 8 '15 at 11:55
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You are counting certain combinations of questions more than once. This is not so easy to see for your example, thus, let's simplify it to a trivial example: you have to choose at least 5 questions out of 5 of part A (i.e. all of A), and at least 1 out of 7 of part B, and, as before, you have to choose 8 questions in total.

We can trivially see that we indeed have to choose 3 questions out of B to satisfy both conditions, since we have already chosen all of A. Thus, the correct answer would be $\left(\begin{matrix}5\\5\end{matrix}\right)\left(\begin{matrix}7\\3\end{matrix}\right)=1\cdot 35=35$.

Your way of solving this question would however say that there are $\left(\begin{matrix}5\\5\end{matrix}\right)\left(\begin{matrix}7\\1\end{matrix}\right)\left(\begin{matrix}6\\2\end{matrix}\right)=1\cdot 7\cdot 15=105$ possibilities of choosing.

Edit: to clarify the problem in this simplified example: assume you choose questions 1,2 and 3 from B. In your calculations there are three possibilities to do so: you choose question 1 because you have to choose at least one from B (Condition 1), and you choose questions 2 and 3 because you have to choose in total 8 (Condition 2). Or you choose question 2 to satisfy Condition 1, and questions 1 and 3 for Condition 2. Finally, you can choose question 3 to satisfy Condition 1, and questions 1 and 2 for Condition 2. In other words, you count each combination 3 times, and, indeed $35\cdot3=105$.

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