There are $10$ persons named $P_1, P_2, P_3, \ldots P_{10}$. Out of $10$ persons, $5$ persons are to be arranged in a line such that in each arrangement $P_1$ must occur whereas $P_4$ and $P_5$ do not occur. Find the number of such possible arrangements.

There are 10 persons named P₁, P₂, P₃, … P₁₀. Out of 10 persons, 5 persons are to be arranged in a line such 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 must choose and arrange 5 people from 10, with $P_1$ mandatory and $P_4$ , $P_5$ forbidden. Since $P_1$ is fixed to be included, we select 4 more from the 7 allowed people and arrange all 5, giving $\binom{7}{4} \times 5! = 35 \times 120 = \boxed{4200}$ arrangements.

Understanding the constraint structure

We're arranging 5 people in a line from a pool of 10, but with restrictions on who can appear. Think of this as a two-stage process: first select which 5 people will participate, then arrange them in order.

The constraints tell us:

$P_1$ must be in every arrangement (mandatory inclusion)
$P_4$ and $P_5$ cannot be in any arrangement (forbidden)
The remaining people $P_2, P_3, P_6, P_7, P_8, P_9, P_{10}$ are available (7 people)

Since $P_1$ is already taking one of the 5 spots, we need to fill the remaining 4 spots from the 7 available people.

Step-by-step solution

Identify the available pool after applying constraints.

We have 10 people total. Removing $P_4$ and $P_5$ leaves us with 8 people: $\{P_1, P_2, P_3, P_6, P_7, P_8, P_9, P_{10}\}$ .

But $P_1$ is mandatory, so we've already "used" one spot. We need 4 more people from the remaining 7: $\{P_2, P_3, P_6, P_7, P_8, P_9, P_{10}\}$ .
Choose 4 people from the 7 available.

The number of ways to select 4 people from 7 is the combination:

\binom{7}{4} = \frac{7!}{4! \cdot 3!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35

Arrange the 5 selected people (including $P_1$ ) in a line.

Once we've chosen which 4 people join $P_1$ , we have 5 people total to arrange in a line. The number of permutations of 5 distinct people is:

5! = 120

Apply the multiplication principle. …

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Understanding the constraint structure

Step-by-step solution

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