Question

In an examination, a student has to answer 4 questions out of 5 questions; questions 1 and 2 are however compulsory. Determine the number of ways in which the student can make the choice.

Answer 1

Hint: This problem can be solved by using the Combinations. It is a way of selecting r objects out of n (arrangement does not matter). It is given by $${n_{{C_r}}} = \dfrac{{{\text{n}}!}}{{\left( {{\text{n - r}}} \right)! \times {\text{r!}}}}$$. From the given problem $n$ here is 3 and $r$ is 2. Therefore the number of ways in which the student can make a choice is ${3_{{C_2}}}$.

Complete step by step solution:
Given:
In an examination, a student has to answer 4 questions.
Therefore,
Total number of question to be answered = 4
Compulsory questions to be answered = 1 and 2
The compulsory questions now consider as a one quantity
Therefore the number of items n = 5-2 = 3
And the number of items to choose at a time r = 4-2 = 2
Now to calculate the number of favorable outcomes, we need to use the combination formula$${n_{{C_r}}} = \dfrac{{{\text{n}}!}}{{\left( {{\text{n - r}}} \right)! \times {\text{r!}}}}$$, where n represents the number of items and r represents the number of items being chosen at a time. Combinations are the way of selecting the objects or numbers from a group of objects or collection, in such a way that the order of the objects does not matter.
Hence the number of ways a student can make choice = ${3_{{C_2}}}$
$ = \dfrac{{3!}}{{\left( {3 - 2} \right)! \times 2!}}$
$ \Rightarrow \dfrac{{3 \times 2 \times 1}}{{2 \times 1}}$
= 3 ways
Therefore, the number of ways in which the student can make the choice is 3 ways.

Note: There are different types of problems that can be solved by these Permutations and combinations. Arrangement is n items can be arranged in $${\text{n!}}$$ways. Permutation is a way of selecting and arranging r objects out of a set of n objects, $${n_{{P_r}}} = \dfrac{{{\text{n}}!}}{{\left( {{\text{n - r}}} \right)!}}$$