Question

In a monthly test, the teacher decides that there will be three questions, one from each of exercise 5, 6 and 7 of the textbook. If there are 12 questions in exercise 5, 18 in exercise 6 and 9 in exercise 7, in how many ways can three questions be selected?

Answer 1

We have to apply combinations here. The number of ways to choose r items from n items is given by $^{n}{{C}_{r}}$ . We have to find the number of ways of choosing 1 question from 12 questions in exercise 5, 1 question from 18 questions in exercise 6 and 9 questions in exercise 7. We have to multiply these values to get the required solution.

Complete step by step answer:
We have to find the number of ways in which 3 questions can be selected from exercises 5, 6, and 7 (one from each exercise).
We are given that exercise 5 has 12 questions. We know that the number of ways to choose r items from n items is given by $^{n}{{C}_{r}}$ . Therefore, the number of ways to choose 1 question from 12 questions is given by
${{\Rightarrow }^{12}}{{C}_{1}}$
We know that $^{n}{{C}_{1}}=n$ .
${{\Rightarrow }^{12}}{{C}_{1}}=12$
We are given that exercise 6 has 18 questions. Therefore, the number of ways to choose 1 question from 18 questions is given by
${{\Rightarrow }^{18}}{{C}_{1}}=18$
We are given that exercise 7 has 9 questions. Therefore, the number of ways to choose 1 question from 9 questions is given by
${{\Rightarrow }^{9}}{{C}_{1}}=9$
Therefore, we can find the number of ways of choosing 3 questions by multiplying the number of ways of selecting one question from 5, 6, and 7.
$\Rightarrow$ Required number of ways $=12\times 18\times 9=1944$ ways.
Hence, the number of ways in which 3 questions can be selected each from exercises 5, 6, and 7 is 1944.

Note: Students must be thorough with the formulas and properties of combinations. They have a chance of making a mistake by writing the property $^{n}{{C}_{1}}$ as n. The formula of combination is given by $^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$ . Students have a chance of making a mistake when finding the required number of ways by adding the number of ways of selecting one question from 5, 6, and 7. They must note that we used combination here instead of permutation because permutations are used when order is concerned.