Eight chairs are numbered $1$ to $8$. Two women and $3$ men wish to occupy one chair each. First the women choose the chairs from amongst the chairs $1$ to $4$ and then men select from the remaining chairs. Find the total number of possible arrangements.

Eight chairs are numbered 1 to 8. Two women and 3 men wish to occupy one chair each. First the women choose th

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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

Watch out

Do not use the permutation formula when order doesn't matter. For example, choosing 3 friends from a group of 5 to form a committee — here the order of selection is irrelevant. That's a combination, not a permutation. Permutations are for ordered arrangements (like rankings, seating orders, passwords where position matters).

Quick Examples

Scenario	Calculation	Answer
Arranging 4 different trophies on a shelf	$4!$	24
Number of 3-digit codes from digits 1–9 (no digit repeated)	$P(9,3) = 9 \times 8 \times 7$	504
Seating 5 people in 5 chairs	$5!$	120
Assigning gold, silver, bronze medals to 8 runners	$P(8,3) = 8 \times 7 \times 6$	336

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.

The problem is a two‑stage selection‑and‑arrangement: first the two women choose 2 distinct chairs from chairs 1–4 (order matters because they are distinct people), then the three men arrange themselves in 3 of the remaining 6 chairs. The total number of arrangements is $4 \times 3 \times 6 \times 5 \times 4 = 1440$ .

The key idea here is that the women and men do not choose chairs simultaneously. The women pick first, and only from chairs 1 to 4. After they have taken their seats, the men pick from whatever chairs are left — which could be anywhere from 1 to 8, except the two already taken.

Because the people are distinct (each woman is a different person, each man is a different person), the order in which they occupy chairs matters. This is a permutation without repetition problem: we are arranging distinct people into distinct chairs, but with a restriction on which chairs the women may initially choose from.

Let’s break it into the two clear stages.

Women choose and sit in chairs 1–4

There are 4 chairs available (numbered 1, 2, 3, 4). Two distinct women need to pick two different chairs.

The first woman has 4 choices. After she sits, the second woman has 3 remaining chairs to choose from.

So the number of ways for the women to occupy two chairs from the set $\{1,2,3,4\}$ is:

4 \times 3 = 12

This is a permutation: $P(4,2) = \frac{4!}{(4-2)!} = 12$ .

Tip

If the women were identical, we would use combinations ( $\binom{4}{2}=6$ ). But since they are different people, swapping them gives a different arrangement — so we multiply, not divide.

Men choose from the remaining chairs

After the two women have taken their seats, there are $8 - 2 = 6$ chairs left. Three distinct men need to sit in three of these chairs.

The first man has 6 choices, the second has 5, the third has 4. So the number of ways for the men is:

6 \times 5 \times 4 = 120

That is $P(6,3) = \frac{6!}{3!} = 120$ .

Combine the two stages

Since the women’s choice and the men’s choice are independent (the men’s options depend on which chairs the women took, but the counting already accounts for all possibilities), we multiply:

12 \times 120 = 1440

Watch out

A common mistake is to treat the women’s choice as $\binom{4}{2}=6$ and then multiply by $5!$ for the men, forgetting that the women are distinct. That gives $6 \times 120 = 720$ , which is exactly half the correct answer. Always check: are the people identical or distinct? Here, they are distinct individuals.

✓Final answer

\boxed{1440}

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Q.Eight chairs are numbered 111 to 888. Two women and 333 men wish to occupy one chair each. First the women choose the chairs from amongst the chairs 111 to 444 and then men select from the remaining chairs. Find the total number of possible arrangements.

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Q.Eight chairs are numbered $1$ to $8$ . Two women and $3$ men wish to occupy one chair each. First the women choose the chairs from amongst the chairs $1$ to $4$ and then men select from the remaining chairs. Find the total number of possible arrangements.

Unlock everything free for 14 days