Find the number of positive integers greater than 6000 and less than 7000 which are divisible by 5, provided t

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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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We need to form 4-digit numbers between 6000 and 7000, divisible by 5, with no repeated digits. This means the thousands digit is 6, and the units digit is either 0 or 5. By considering these two cases and counting the available choices for the remaining digits, we find there are 112 such integers.

To find the number of positive integers that meet all the given conditions, we will determine the number of choices for each digit position, moving from the most constrained positions first. The number must be a 4-digit number, say $d_1 d_2 d_3 d_4$ , where $d_1$ is the thousands digit, $d_2$ is the hundreds digit, $d_3$ is the tens digit, and $d_4$ is the units digit.

Here are the conditions we must satisfy:

The number is greater than $6000$ and less than $7000$ .
The number is divisible by $5$ .
No digit is to be repeated.

Let's break this down step by step.

Determine the thousands digit ( $d_1$ ).

The condition that the number is greater than $6000$ and less than $7000$ immediately tells us that it must be a 4-digit number, and its thousands digit must be $6$ .

So, $d_1 = 6$ .

There is only $1$ choice for the thousands digit.

Digits available for use: $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$ .

Digit used: $6$ .
Determine the units digit ( $d_4$ ).

For a number to be divisible by $5$ , its units digit must be either $0$ or $5$ .

So, $d_4 \in \{0, 5\}$ .

Since the thousands digit $d_1 = 6$ is neither $0$ nor $5$ , there is no immediate conflict with the "no repeated digits" rule for $d_1$ and $d_4$ . However, the choice of $d_4$ will affect the remaining available digits for $d_2$ and $d_3$ . We will consider two separate cases based on the value of $d_4$ .
Case 1: The units digit ( $d_4$ ) is $0$ .
- Thousands digit $d_1 = 6$ (1 choice).
- Units digit $d_4 = 0$ (1 choice).
- Digits used so far: $6$ and $0$ .
- Remaining available digits for $d_2$ and $d_3$ : From the set $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$ , we exclude $6$ and $0$ . This leaves us with $10 - 2 = 8$ digits: $\{1, 2, 3, 4, 5, 7, 8, 9\}$ .
- Determine the hundreds digit ( $d_2$ ). We can choose any of the $8$ remaining available digits for $d_2$ . So, there are $8$ choices for $d_2$ .
- Determine the tens digit ( $d_3$ ). After choosing $d_2$ , we have used three distinct digits ( $d_1, d_4, d_2$ ). There are $10 - 3 = 7$ digits remaining for $d_3$ . So, there are $7$ choices for $d_3$ .
- The number of integers in this case is $1 \times 8 \times 7 \times 1 = 56$ .
Case 2: The units digit ( $d_4$ ) is $5$ .
- Thousands digit $d_1 = 6$ (1 choice). …

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Q.Find the number of positive integers greater than 600060006000 and less than 700070007000 which are divisible by 555, provided that no digit is to be repeated.

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Q.Find the number of positive integers greater than $6000$ and less than $7000$ which are divisible by $5$ , provided that no digit is to be repeated.

Unlock everything free for 14 days