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Question: A candidate is required to answer 6 questions out of 10 questions which are divided into two groups ...

A candidate is required to answer 6 questions out of 10 questions which are divided into two groups each containing 5 questions and he is not permitted to attempt more than 4 from each group. In how many ways can he make up his choice?

Explanation

Solution

Hint: Consider all possible positive integral solutions of the equation x+y=6x+y=6 such that x4,y4x\le 4,y\le 4. Here x and y represent the number of questions that a person can choose from each set. Count the number of ways to choose x and y questions from a set of 5 questions each. Multiply the value of the number of ways to choose 6 questions for a fixed value of x and y. Add all the ways to choose 6 questions for all possible values of x and y.

Complete step-by-step solution -
We have to calculate the number of ways in which a candidate can attempt 6 questions out of 10 questions which are divided into two groups, each containing 5 questions such that the candidate can attempt at most 4 questions from each group.
Let’s assume that the candidate chooses x questions from the first set and y questions from the second set. We have x+y=6x+y=6 such that x4,y4x\le 4,y\le 4.
We will find all possible values of x and y which satisfy the above equation.
For x=4x=4, we have y=2y=2.
For x=3x=3, we have y=3y=3.
For x=2x=2, we have y=4y=4.
These are the only possible ways to choose the number of questions to be attempted from each set.
We know that we can choose r objects from a set of n objects in nCr{}^{n}{{C}_{r}} ways.
We will now find the number of ways to choose questions from each given set.
For x=4,y=2x=4,y=2, number of ways to choose 4 questions from a set of 5 questions is 5C4=5!4!1!=5{}^{5}{{C}_{4}}=\dfrac{5!}{4!1!}=5 and number of ways to choose 2 questions from a set of 5 questions is 5C2=5!2!3!=10{}^{5}{{C}_{2}}=\dfrac{5!}{2!3!}=10. So, the number of ways to choose 6 questions out of 10 questions such that 4 questions are chosen from the first set and 2 are chosen from the second set =5×10=50=5\times 10=50.
For x=3,y=3x=3,y=3, number of ways to choose 3 questions from a set of 5 questions is 5C3=5!3!2!=10{}^{5}{{C}_{3}}=\dfrac{5!}{3!2!}=10 and number of ways to choose 3 questions from a set of 5 questions is 5C3=5!2!3!=10{}^{5}{{C}_{3}}=\dfrac{5!}{2!3!}=10. So, the number of ways to choose 6 questions out of 10 questions such that 3 questions are chosen from the first set and 3 are chosen from the second set =10×10=100=10\times 10=100.
For x=2,y=4x=2,y=4, number of ways to choose 4 questions from a set of 5 questions is 5C4=5!4!1!=5{}^{5}{{C}_{4}}=\dfrac{5!}{4!1!}=5 and number of ways to choose 2 questions from a set of 5 questions is 5C2=5!2!3!=10{}^{5}{{C}_{2}}=\dfrac{5!}{2!3!}=10. So, the number of ways to choose 6 questions out of 10 questions such that 2 questions are chosen from the first set and 4 are chosen from the second set =5×10=50=5\times 10=50.
So, the total number of ways to choose 6 questions from a set of 10 questions is the sum of all possible ways to choose 6 questions, which is =50+100+50=200=50+100+50=200.
Hence, there are 200 ways to choose 6 questions from a set of 10 questions which are divided into two sets having 5 questions each.

Note: One must keep in mind that we are not only supposed to choose the number of questions to be picked from each set; we also have to count the ways to choose the questions. Also, one must know that nCr=n!r!(nr)!{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}.