Q.How many numbers lying between 100 and 1000 can be formed with the digits 0, 1, 2, 3, 4, 5, if the repetition of the digits is not allowed?

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 $3!$.

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 ${}^5P_3$.

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

Common Mistake to Avoid

Quick Examples

The Bottom Line

Permutations without repetition answer the question: "In how many different ordered ways can I arrange a set of distinct items, using each item at most once?" The answer is always a product of decreasing integers, starting from $n$ and going down $r$ steps. When $r = n$, it's simply $n!$.

Permutations Without Repetition is introduced in the NCERT Class 11 Mathematics Permutations and Combinations chapter, and it is exactly the kind of topic students look up when searching "permutations formula class 11 maths" or "arrangement of distinct objects important questions". It also forms the basis for many JEE Main and state CET counting problems that ask you to arrange distinct items without repeating any of them.