Question

A candidate is required to answer 7 questions out of 12 questions which are divided into two groups each containing 6 questions. He is not permitted to attempt more than 5 questions from each group. The number of ways in which he can choose the 7 questions is:

Answer 1

Hint: Here, a student can do only 7 questions such that he cannot choose more than 5 questions from any group. So find the number of ways when he chooses 5 questions from ${{S}_{1}}$ and 2 questions from ${{S}_{2}}$ and vice versa. And also, when he chooses 3 questions from ${{S}_{1}}$ and 4 questions from ${{S}_{2}}$ and vice versa. Here ${{S}_{1}}$ and ${{S}_{2}}$ are two groups.

Complete step-by-step answer:

Here, we are given that a candidate is required to answer 7 out of 12 questions which are divided into two groups of 6 questions. Also, he is not permitted to attempt more than 5 questions from each group. We have to find the number of ways in which he can choose 7 questions.
Let us consider two groups as ${{S}_{1}}$ and ${{S}_{2}}$.
We know that both groups ${{S}_{1}}$ and ${{S}_{2}}$ have 6 questions each. Therefore the total number of questions is 12.
We are given that a student is required to do 7 out of 12 questions and he is not permitted to do more than 5 questions from each group. So he can choose 7 questions in these ways:
1. He can choose 5 questions from ${{S}_{1}}$ and 2 questions from ${{S}_{2}}$.
2. He can choose 4 questions from ${{S}_{1}}$ and 3 questions from ${{S}_{2}}$.
3. He can choose 3 questions from ${{S}_{1}}$ and 4 questions from ${{S}_{2}}$.
4. He can choose 2 questions from ${{S}_{1}}$ and 5 questions from ${{S}_{2}}$.
So, we get,
1. Number of ways of choosing 5 questions out of 6 questions from ${{S}_{1}}$ and 2 questions out of 6 questions from ${{S}_{2}}$

& =6{{C}_{5}}\times 6{{C}_{2}} \\\ & =\dfrac{6!}{5!\left( 6-5 \right)!}\times \dfrac{6!}{2!\left( 6-2 \right)!} \\\ \end{aligned}$$ $$\begin{aligned} & =\dfrac{6!}{5!}\times \dfrac{6!}{2!\times 4!} \\\ & =\dfrac{6\times 6\times 5}{2} \\\ & =90\text{ ways} \\\ \end{aligned}$$ 2\. Similarly the number of ways of choosing 4 questions from $${{S}_{1}}$$ and 3 questions from $${{S}_{2}}$$ $$\begin{aligned} & =6{{C}_{4}}\times 6{{C}_{3}} \\\ & =\dfrac{6!}{4!2!}\times \dfrac{6!}{3!3!} \\\ \end{aligned}$$ $$\begin{aligned} & =\dfrac{\left( 6\times 5 \right)\left( 6\times 5\times 4 \right)}{2\times 6} \\\ & =300\text{ ways} \\\ \end{aligned}$$ 3\. Similarly the number of ways of choosing 3 questions from $${{S}_{1}}$$ and 4 questions from $${{S}_{2}}$$ $$\begin{aligned} & =6{{C}_{3}}\times 6{{C}_{4}} \\\ & =\dfrac{6!}{3!3!}\times \dfrac{6!}{4!2!} \\\ \end{aligned}$$ $$\begin{aligned} & =\dfrac{\left( 6\times 5\times 4 \right)\left( 6\times 5 \right)}{2\times 6} \\\ & =300\text{ ways} \\\ \end{aligned}$$ 4\. Similarly the number of ways of choosing 2 questions from $${{S}_{1}}$$ and 5 questions from $${{S}_{2}}$$ $$\begin{aligned} & =6{{C}_{2}}\times 6{{C}_{5}} \\\ & =\dfrac{6!}{2!4!}\times \dfrac{6!}{5!1!} \\\ \end{aligned}$$ $$\begin{aligned} & =\dfrac{\left( 6\times 5 \right)\times \left( 6 \right)}{2} \\\ & =90\text{ ways} \\\ \end{aligned}$$ So, we get the total number of ways in which he can choose 7 questions = 90 + 300 + 300 + 90 = 780 ways. Note: In this question, many students consider case 1 and case 4 or case 2 and case 3 as similar cases due to similar values and take them as one case and arrive at the wrong answer. But they must note that both cases are different even if the values are the same because groups 1 and 2 are different and they will contain different questions. Also, take care that in every way, both conditions that are given in questions are satisfied and recheck the calculations to avoid any mistakes.