To fill $12$ vacancies there are $25$ candidates of which $5$ are from scheduled castes. If $3$ of the vacancies are reserved for scheduled caste candidates while the rest are open to all, the number of ways in which the selection can be made is ${}^{5}C_{3} \times {}^{20}C_{9}$.

> [!TLDR] > The selection process is divided into two independent parts: choosing 3 Scheduled Caste (SC) candidates for reserved seats and choosing 9 non-SC candidates for open seats. The total number of ways is the product of the combinations for each part, resulting in $\boxed{{}^{5}C_{3} \times {}^{20}C_{9}}$. The problem asks us to find the total number of ways to select candidates for a set of vacancies, which are divided into two categories: reserved and open. The core idea here is that these two selection processes are independent of each other. We first select candidates for the reserved seats, and then we select candidates for the open seats. Since the order in which candidates are chosen does not matter, this is a problem of combinations. We will break down the selection into two distinct stages and then multiply the number of ways for each stage to get the total number of ways. 1. **Identify the categories of vacancies and candidates:** We have a total of $12$ vacancies. * $3$ vacancies are reserved for Scheduled Caste (SC) candidates. * The remaining $12 - 3 = 9$ vacancies are open to all. We have a total of $25$ candidates. * $5$ candidates are from Scheduled Castes (SC). * The remaining $25 - 5 = 20$ candidates are non-SC. The problem statement provides the answer as ${}^{5}C_{3} \times {}^{20}C_{9}$. This implies a specific interpretation for the "open to all" vacancies: that these $9$ vacancies are filled *only* from the pool of non-SC candidates. This effectively creates two completely separate selection pools. 2. **Selection for the Reserved Vacancies:** We need to fill $3$ reserved vacancies, and these must be filled by SC candidates. We have $5$ SC candidates available. The number of ways to choose $3$ SC candidates from $5$ SC candidates is given by the combination formula: > [!FORMULA] > The number of combinations of $n$ distinct items taken $r$ at a time is given by: > $${}^{n}C_r

To fill 12 vacancies there are 25 candidates of which 5 are from scheduled castes. If 3 of the vacancies are r

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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 selection process is divided into two independent parts: choosing 3 Scheduled Caste (SC) candidates for reserved seats and choosing 9 non-SC candidates for open seats. The total number of ways is the product of the combinations for each part, resulting in $\boxed{{}^{5}C_{3} \times {}^{20}C_{9}}$ .

The problem asks us to find the total number of ways to select candidates for a set of vacancies, which are divided into two categories: reserved and open. The core idea here is that these two selection processes are independent of each other. We first select candidates for the reserved seats, and then we select candidates for the open seats. Since the order in which candidates are chosen does not matter, this is a problem of combinations.

We will break down the selection into two distinct stages and then multiply the number of ways for each stage to get the total number of ways.

Identify the categories of vacancies and candidates:

We have a total of $12$ vacancies.
- $3$ vacancies are reserved for Scheduled Caste (SC) candidates.
- The remaining $12 - 3 = 9$ vacancies are open to all.
We have a total of $25$ candidates.
- $5$ candidates are from Scheduled Castes (SC).
- The remaining $25 - 5 = 20$ candidates are non-SC.
The problem statement provides the answer as ${}^{5}C_{3} \times {}^{20}C_{9}$ . This implies a specific interpretation for the "open to all" vacancies: that these $9$ vacancies are filled only from the pool of non-SC candidates. This effectively creates two completely separate selection pools.
Selection for the Reserved Vacancies:

We need to fill $3$ reserved vacancies, and these must be filled by SC candidates. We have $5$ SC candidates available.

The number of ways to choose $3$ SC candidates from $5$ SC candidates is given by the combination formula:
$n$
Selection for the Open Vacancies:

We need to fill $9$ open vacancies. According to the structure of the given answer, these $9$ vacancies are to be filled from the non-SC candidates. We have $20$ non-SC candidates available.

The number of ways to choose $9$ candidates from $20$ non-SC candidates is given by the combination formula:

Here, $n=20$ (total non-SC candidates) and $r=9$ (candidates to be selected for open vacancies).

Number of ways to fill open vacancies $= {}^{20}C_9$ . …

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