Find the number of permutations of n distinct things taken r together, in which 3 particular things must occur

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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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Treat the 3 particular things as a single "super-object," select the remaining $(r-3)$ objects from the other $(n-3)$ things, arrange all units, then arrange internally. The answer is $\boxed{^{n-3}P_{r-3} \times r! \div (r-2)! = \frac{(n-3)! \cdot r!}{(n-r)! \cdot (r-2)!}}$ or equivalently $\boxed{3! \times (r-2)! \times \binom{n-3}{r-3}}$ .

Why this approach works

When certain objects must appear together in a permutation, the standard trick is to bundle them into a single unit. This transforms the problem: instead of arranging $r$ individual objects with a constraint, we arrange fewer objects (the bundle plus others) without constraint, then account for the internal arrangements within the bundle.

The key insight: if 3 things must be together, they occupy 3 consecutive positions in any valid arrangement. By treating them as one "block," we ensure they stay adjacent, then multiply by the ways to arrange them internally.

Step-by-step solution

Identify what we're choosing and arranging

We need $r$ positions filled from $n$ distinct things, with 3 specific things (call them $A$ , $B$ , $C$ ) always together. Since these 3 must occur, we need $r \geq 3$ .
Bundle the 3 particular things

Treat $A$ , $B$ , $C$ as a single composite object $[ABC]$ . Now instead of selecting $r$ things from $n$ , we select:
- The bundle $[ABC]$ (mandatory, takes up 3 of our $r$ slots)
- $(r - 3)$ additional things from the remaining $(n - 3)$ objects
Select the remaining objects

From the $(n-3)$ things (excluding $A$ , $B$ , $C$ ), choose $(r-3)$ in order. The number of ways is:

^{n-3}P_{r-3} = \frac{(n-3)!}{(n-3-(r-3))!} = \frac{(n-3)!}{(n-r)!}

Arrange the "units" We now have $(r-3) + 1 = (r-2)$ objects to arrange: the bundle $[ABC]$ plus $(r-3)$ individual things. These $(r-2)$ units can be permuted in:

(r-2)!

ways.

Arrange internally within the bundle The 3 things inside $[ABC]$ can be arranged among themselves in:

3! = 6

ways. …

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Q.Find the number of permutations of nnn distinct things taken rrr together, in which 333 particular things must occur together.

Why this approach works

Step-by-step solution

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Q.Find the number of permutations of $n$ distinct things taken $r$ together, in which $3$ particular things must occur together.

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