Question

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?

Answer 1

Hint: Consider all possible positive integral solutions of the equation \[x+y=6\] such that \[x\le 4,y\le 4\]. 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 \[x+y=6\] such that \[x\le 4,y\le 4\].
We will find all possible values of x and y which satisfy the above equation.
For \[x=4\], we have \[y=2\].
For \[x=3\], we have \[y=3\].
For \[x=2\], we have \[y=4\].
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 \[{}^{n}{{C}_{r}}\] ways.
We will now find the number of ways to choose questions from each given set.
For \[x=4,y=2\], number of ways to choose 4 questions from a set of 5 questions is \[{}^{5}{{C}_{4}}=\dfrac{5!}{4!1!}=5\] and number of ways to choose 2 questions from a set of 5 questions is \[{}^{5}{{C}_{2}}=\dfrac{5!}{2!3!}=10\]. 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 \[=5\times 10=50\].
For \[x=3,y=3\], number of ways to choose 3 questions from a set of 5 questions is \[{}^{5}{{C}_{3}}=\dfrac{5!}{3!2!}=10\] and number of ways to choose 3 questions from a set of 5 questions is \[{}^{5}{{C}_{3}}=\dfrac{5!}{2!3!}=10\]. 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 \[=10\times 10=100\].
For \[x=2,y=4\], number of ways to choose 4 questions from a set of 5 questions is \[{}^{5}{{C}_{4}}=\dfrac{5!}{4!1!}=5\] and number of ways to choose 2 questions from a set of 5 questions is \[{}^{5}{{C}_{2}}=\dfrac{5!}{2!3!}=10\]. 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 \[=5\times 10=50\].
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 \[=50+100+50=200\].
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 \[{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\].