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.

> [!TLDR] > The student must answer 4 questions from a set of 5, with questions 1 and 2 compulsory. This means the student must choose the remaining 2 questions from the 3 optional ones (questions 3, 4, 5). The number of ways is simply the number of combinations of 3 items taken 2 at a time, which is $\binom{3}{2} = 3$. The core idea here is **selection without regard to order**. The student isn't arranging the questions in a sequence; they are simply picking which ones to answer. Since questions 1 and 2 are forced, the only freedom lies in choosing the remaining questions from the pool of optional ones. Let’s break it down step by step. 1. **Identify the compulsory set.** The problem states that questions 1 and 2 are compulsory. That means the student *must* include both of these in their set of 4 answers. So, 2 out of the 4 required questions are already fixed. 2. **Determine what remains to be chosen.** The student needs a total of 4 questions. They already have 2 (questions 1 and 2). So they need to pick $4 - 2 = 2$ more questions from the remaining ones. 3. **List the available options.** The total questions are 5: numbered 1, 2, 3, 4, 5. After removing the compulsory 1 and 2, the remaining questions are 3, 4, and 5. That’s a set of 3 questions. 4. **Apply the combination formula.** The student must choose 2 questions from these 3. The order in which they pick them doesn’t matter — choosing {3, 4} is the same as choosing {4, 3}. So we use combinations, not permutations. The number of ways to choose $r$ items from $n$ distinct items is given by the binomial coefficient: $$ \binom{n}{r} = \frac{n!}{r!(n-r)!} $$ Here, $n = 3$ and $r = 2$, so: $$ \binom{3}{2} = \frac{3!}{2! \cdot 1!} = \frac{3 \times 2 \times 1}{(2 \times 1) \times 1} = 3 $$ 5. **Interpret the result.** …

In an examination, a student has to answer 4 questions out of 5 questions; questions 1 and 2 are however compu

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Concept understanding — Permutations Without Repetition

Permutations Without Repetition – The Idea of Arranging Things

Imagine you have three different books on a shelf: a Physics book, a Chemistry book, and a Maths book. How many different ways can you arrange them in a row?

You could try listing them out:

Physics, Chemistry, Maths
Physics, Maths, Chemistry
Chemistry, Physics, Maths
Chemistry, Maths, Physics
Maths, Physics, Chemistry
Maths, Chemistry, Physics

That's 6 arrangements. Notice that each arrangement uses all three books exactly once — no book is repeated, and no book is left out. This is the core idea: permutations without repetition count the number of ways to arrange a set of distinct objects in order, using each object exactly once.

Why "Without Repetition"?

The phrase "without repetition" means that once you place an object in a position, you cannot use it again. In our book example, once you put the Physics book in the first slot, you cannot put it in the second or third slot. Each object appears exactly once in the arrangement.

This is different from "permutations with repetition" (like creating 3-letter codes from the letters A, B, C where you can reuse letters — e.g., AAA, AAB, etc.). Here, no repeats allowed.

The Counting Logic – Why Multiply?

Let's build the arrangement step by step for 3 distinct books:

First position: You have 3 choices (any of the 3 books).
Second position: After placing the first book, only 2 books remain — so 2 choices.
Third position: Only 1 book is left — so 1 choice.

Total arrangements = $3 \times 2 \times 1 = 6$ .

This product $3 \times 2 \times 1$ is called 3 factorial, written as $3!$ .

P(n) = n! = n \times (n-1) \times (n-2) \times \cdots \times 2 \times 1

For $n$ distinct objects, the number of permutations (arrangements in order) is $n!$ .

What If You Only Arrange Some of Them?

Suppose you have 5 different books, but you only want to arrange 3 of them on a shelf. How many ways?

First position: 5 choices
Second position: 4 choices
Third position: 3 choices

Total = $5 \times 4 \times 3 = 60$ .

This is a permutation of 5 objects taken 3 at a time, written as $P(5,3)$ or ${}^5P_3$ .

P(n, r) = \frac{n!}{(n-r)!} = n \times (n-1) \times \cdots \times (n-r+1)

Here $n$ is the total number of distinct objects, and $r$ is how many you are arranging. The formula works because:

Numerator $n!$ counts all arrangements of all $n$ objects.
Denominator $(n-r)!$ removes the arrangements of the $n-r$ objects you are not using.

Key Points to Remember

Important

Order matters — swapping two objects gives a different permutation.
No repetition — each object is used at most once.
For arranging all $n$ objects: $n!$
For arranging r out of $n$ objects: $\frac{n!}{(n-r)!}$

Common Mistake to Avoid …

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The student must answer 4 questions from a set of 5, with questions 1 and 2 compulsory. This means the student must choose the remaining 2 questions from the 3 optional ones (questions 3, 4, 5). The number of ways is simply the number of combinations of 3 items taken 2 at a time, which is $\binom{3}{2} = 3$ .

The core idea here is selection without regard to order. The student isn't arranging the questions in a sequence; they are simply picking which ones to answer. Since questions 1 and 2 are forced, the only freedom lies in choosing the remaining questions from the pool of optional ones.

Let’s break it down step by step.

Identify the compulsory set.

The problem states that questions 1 and 2 are compulsory. That means the student must include both of these in their set of 4 answers. So, 2 out of the 4 required questions are already fixed.
Determine what remains to be chosen.

The student needs a total of 4 questions. They already have 2 (questions 1 and 2). So they need to pick $4 - 2 = 2$ more questions from the remaining ones.
List the available options.

The total questions are 5: numbered 1, 2, 3, 4, 5. After removing the compulsory 1 and 2, the remaining questions are 3, 4, and 5. That’s a set of 3 questions.
Apply the combination formula.

The student must choose 2 questions from these 3. The order in which they pick them doesn’t matter — choosing {3, 4} is the same as choosing {4, 3}. So we use combinations, not permutations.

The number of ways to choose $r$ items from $n$ distinct items is given by the binomial coefficient:

\binom{n}{r} = \frac{n!}{r!(n-r)!}

Here, $n = 3$ and $r = 2$ , so:

\binom{3}{2} = \frac{3!}{2! \cdot 1!} = \frac{3 \times 2 \times 1}{(2 \times 1) \times 1} = 3

Interpret the result. …

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Q.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.

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