The number of permutations of n different objects, taken r at a line, when repetitions are allowed, is ___.

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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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When repetitions are allowed, each of the $r$ positions can be filled independently with any of the $n$ objects, giving $n^r$ permutations. The answer is $n^r$ .

The core idea

Permutations without repetition are about arranging distinct objects — once you use an object, it's gone. But when repetitions are allowed, you have a completely different situation: every time you pick an object for a position, you have the full set of $n$ objects available again. This is like choosing a password or a code where digits or letters can repeat.

Think of it this way: you are filling $r$ slots in a line. For the first slot, you have $n$ choices. For the second slot, because you can reuse the same object, you again have $n$ choices. And so on, for all $r$ slots. The choices are independent — what you pick for one slot doesn't affect the next.

Step-by-step reasoning

Understand the counting principle.

The fundamental principle of counting says: if one event can happen in $m$ ways and another independent event can happen in $n$ ways, then the two events together can happen in $m \times n$ ways. Here, each slot is an independent event.
Apply it to the first slot.

You have $n$ different objects. For the first position in the line, you can choose any one of them.

Number of ways = $n$ .
Apply it to the second slot.

Since repetition is allowed, you can again choose any of the $n$ objects — even the one you just used.

Number of ways = $n$ .
Continue for all $r$ slots.

Each of the $r$ positions has exactly $n$ choices, independent of the others.
Multiply the choices. …

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Q.The number of permutations of nnn different objects, taken rrr at a line, when repetitions are allowed, is ______.

The core idea

Step-by-step reasoning

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Q.The number of permutations of $n$ different objects, taken $r$ at a line, when repetitions are allowed, is ______.

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