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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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Each of the $12$ stalls can be filled with any of $3$ types of animals (horse, cow, or calf), giving $3^{12}$ arrangements by the multiplication principle.

The heart of this problem lies in recognizing that we're making independent choices. We have $12$ distinct stalls (positions matter) and an unlimited supply of three types of animals. Since there are "not less than $12$ each," we never run out of any type—every stall can be filled with any of the three animals without restriction.

Think of it this way: you're standing at stall $1$ and asking, "What can I put here?" You have $3$ choices. Move to stall $2$ —again, $3$ choices, completely independent of what you chose for stall $1$ . This independence is the key.

Why the multiplication principle applies

When we perform a sequence of independent tasks, the total number of ways to complete all tasks is the product of the number of ways to complete each individual task. Here:

Task 1: Fill stall $1$ → $3$ ways
Task 2: Fill stall $2$ → $3$ ways
$\vdots$
Task 12: Fill stall $12$ → $3$ ways

Since each choice is independent (the animals are plentiful), we multiply:

3 \times 3 \times 3 \times \cdots \times 3 \text{ (12 times)} = 3^{12}

Step-by-step reasoning

Identify the positions. We have $12$ stalls, each of which is a distinct position. Stall $1$ is different from stall $2$ , so the arrangement $(H, C, C, \ldots)$ is different from $(C, H, C, \ldots)$ .
Count choices per position. For each stall, we can place:
- a horse, or
- a cow, or
- a calf.
That's $3$ choices per stall.
Check for independence. The problem states there are "not less than $12$ each" of horses, cows, and calves. This means we have at least $12$ of each type available. Even if we filled all $12$ stalls with horses, we'd still have horses left over. So our choice for any stall doesn't restrict our choices for other stalls. …

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Q.In a steamer there are stalls for 121212 animals, and there are horses, cows and calves (not less than 121212 each) ready to be shipped. They can be loaded in 3123^{12}312 ways.

Why the multiplication principle applies

Step-by-step reasoning

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Q.In a steamer there are stalls for $12$ animals, and there are horses, cows and calves (not less than $12$ each) ready to be shipped. They can be loaded in $3^{12}$ ways.

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