Answer

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**Hint:**This question needs the concept of combinations. First analyse the given question and figure out the different cases in which the students can attempt the required number of questions and then find the number of ways in which he can answer in each case.

**Complete step by step solution:**

It is given in the question, that a question paper consists of a number of 10 questions in total. These questions are equally divided into two sections a and b. This means that each section consists of 5 questions. Now, it is said that the student has to attempt six questions in total. However, there is a condition that the student has to attempt at least two questions from each other. Let us analyse and identify the different cases in which the student can attempt the required questions according to the specified condition.

**Case 1:**

The student can attempt 4 questions from section a and 2 questions from section b.

The number of ways in which the student can select m questions from n questions is equal to ${}^{n}{{C}_{m}}$

Therefore, the number of ways in which the student can choose 4 questions from section a is ${}^{5}{{C}_{4}}=\dfrac{5!}{4!(5-4)!}=\dfrac{5!}{4!(1)!}=5$

And the number of ways for the student to choose 2 questions from section b is ${}^{5}{{C}_{2}}=\dfrac{5!}{2!(5-2)!}=\dfrac{5!}{2!(3)!}=\dfrac{5\times 4}{2}=10$

Hence, the number of ways in case 1 is 15.

**Case 2:**

The student can attempt 3 questions from section a and 3 questions from section b.

The number of ways in which the student can do this is ${}^{5}{{C}_{3}}+{}^{5}{{C}_{3}}=\dfrac{5!}{3!(5-3)!}+\dfrac{5!}{3!(5-3)!}=2\times \dfrac{5!}{2!(2)!}=2\times \dfrac{5\times 4\times 3}{3\times 2}=20$

**Case 3:**

The student can attempt 2 questions from section a and 4 questions from section b.

The number of ways in which the student can do this is ${}^{5}{{C}_{2}}+{}^{5}{{C}_{4}}=15$.

**Therefore, the total number of ways in which the students can attempt six questions is $15+20+15=50$.**

**Note:**Some students may make a mistake by multiplying the found answer by two thinking that the same case can be repeated by exchanging sections a and b. In these type of problems we may also use the properties of combinations.In this question we could have used the property ${}^{n}{{C}_{m}}={}^{n}{{C}_{n-m}}$.

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