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Home » 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 either group. Find the number of different ways of doing questions.
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 either group. Find the number of different ways of doing questions.
CBSE | Class 11 | Maths | NCERT Exemplar | Permutations and Combinations
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 either group. Find the number of different ways of doing questions.

Solution:

It is known that,

\begin{array}{l} { }^{n} C_{r} \\ =\frac{n !}{r !(n-r) !} \end{array}

6 = Number of questions in group A

6 = Number of questions in group B

As per the question,

The ways in which we can attempt the questions are,

    \[\begin{tabular}{|l|l|l|l|l|} \hline Group A & 2 & 3 & 4 & 5 \\ \hline Group B & 5 & 4 & 3 & 2 \\ \hline \end{tabular}\]

As a result, the no. of ways of doing questions,

\begin{array}{l} =\left({ }^{6} \mathrm{C}_{2} \times{ }^{6} \mathrm{C}_{5}\right)+\left({ }^{6} \mathrm{C}_{3} \times ^{6} \mathrm{C}_{4}\right)+\left({ }^{6} \mathrm{C}_{4} \times ^{6} \mathrm{C}_{3}\right)+\left({ }^{6} \mathrm{C}_{5} \times ^{6} \mathrm{C}_{2}\right) \\ =(15 \times 6)+(20 \times 15)+(15 \times 20)+(6 \times 15) \\ =780 \end{array}

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