The total number of 9 digit numbers which have all different digits is (A) 10! (B) 9! (C) 9 × 9! (D) 10 × 10!

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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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To form a 9-digit number with all different digits, the first digit cannot be zero, giving 9 choices. The remaining 8 positions can be filled by the remaining 9 distinct digits in $9!$ ways. The total is $\boxed{9 \times 9!}$ .

When we talk about forming "numbers" using digits, two key ideas come into play:

Order matters: A number like $123$ is different from $321$ , even though they use the same digits. This indicates we are dealing with permutations (arrangements), not combinations (selections).
The first digit constraint: For a number to truly be a $k$ -digit number, its first digit (the leftmost, most significant digit) cannot be $0$ . For example, $0123$ is a 3-digit number, not a 4-digit number.

The phrase "all different digits" means that each digit used in the 9-digit number must be unique; no digit can be repeated. This is a classic scenario for permutations without repetition.

Let's break down the process of constructing such a 9-digit number. We have 9 positions to fill, and we need to choose 9 distinct digits from the available 10 digits: $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$ .

Consider the first position (the hundred millions place):

This is the most crucial step. Since we are forming a 9-digit number, the first digit cannot be $0$ .

The available non-zero digits are $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$ .

Therefore, there are $9$ choices for the first digit.

Watch out

A common mistake is to assume $10$ choices for the first digit, forgetting that a number cannot start with $0$ . This would lead to an incorrect result.
Consider the second position:

We have already used one distinct digit for the first position.

Now, we have $10 - 1 = 9$ digits remaining. Importantly, $0$ is now available to be used in this position (or any subsequent position).

So, there are $9$ choices for the second digit.
Consider the third position:

We have used two distinct digits (one for the first position, one for the second).

Now, we have $10 - 2 = 8$ digits remaining.

So, there are $8$ choices for the third digit.
Continue this pattern for the remaining positions:
- For the fourth position, there are $7$ choices.
- For the fifth position, there are $6$ choices.
- For the sixth position, there are $5$ choices.
- For the seventh position, there are $4$ choices.
- For the eighth position, there are $3$ choices.
- For the ninth position, there are $2$ choices. …

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Q.The total number of 999 digit numbers which have all different digits is (A) 10!10!10! (B) 9!9!9! (C) 9×9!9 \times 9!9×9! (D) 10×10!10 \times 10!10×10!

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Q.The total number of $9$ digit numbers which have all different digits is
(A) $10!$
(B) $9!$
(C) $9 \times 9!$
(D) $10 \times 10!$

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