In the permutations of n things, r taken together, the number of permutations in which m particular things occ

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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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When $m$ particular things must occur together in a permutation of $r$ items chosen from $n$ , we treat the $m$ things as a single block. This leads to the correct formula: $\boxed{m! \times (r-m+1) \times {}^{n-m}P_{r-m}}$ . The formula provided in the question, ${}^{n-m}P_{r-m} \times {}^{r}P_{m}$ , is generally incorrect.

Permutations deal with the arrangement of items where the order matters. When we are asked to find the number of permutations where certain items always occur together, the fundamental concept is to treat those items as a single, inseparable unit or "block". This simplifies the problem into two main parts:

Arranging the items within this block.
Arranging this block along with the other individual items.

Let's derive the correct formula for the number of permutations of $n$ distinct things, $r$ taken together, such that $m$ particular things always occur together.

Identify the 'block' of particular things:

We have $m$ particular things that must always occur together. Let's consider these $m$ things as a single unit or a 'block'.
Arrange the items within the block:

The $m$ particular things within their block can be arranged among themselves in $m!$ ways. For example, if the particular things are A, B, C, they can be arranged as ABC, ACB, BAC, BCA, CAB, CBA within their block.

The number of internal arrangements is $m!$ .
Identify the remaining items:

Out of the total $n$ things, $m$ particular things are now grouped into a block. This leaves us with $n-m$ other individual things.
Select the remaining items to form the permutation:

We need to form a permutation of $r$ things. Since the block of $m$ particular things must always be included, it accounts for $m$ of the $r$ positions. Therefore, we need to select an additional $r-m$ things from the remaining $n-m$ individual things.

The number of ways to choose these $r-m$ things from the $n-m$ available things is given by the combination formula: ${}^{n-m}C_{r-m}$ .
Arrange the units (the block and the selected individual items):

Now we have one block (containing the $m$ particular things) and $r-m$ individual things (chosen in the previous step). In total, we have $1 + (r-m)$ units to arrange.

These $(r-m+1)$ units can be arranged in $(r-m+1)!$ ways.
Combine the arrangements:

To find the total number of permutations, we multiply the number of ways to arrange items within the block, the number of ways to choose the remaining items, and the number of ways to arrange all the units.

Total permutations $= (\text{internal arrangements}) \times (\text{ways to choose remaining items}) \times (\text{arrangements of units})$

Total permutations $= m! \times {}^{n-m}C_{r-m} \times (r-m+1)!$
Simplify the expression:

Recall that ${}^{n-m}C_{r-m} = \frac{(n-m)!}{(r-m)!(n-m-(r-m))!} = \frac{(n-m)!}{(r-m)!(n-r)!}$ .

Substituting this into our expression:

Total permutations $= m! \times \frac{(n-m)!}{(r-m)!(n-r)!} \times (r-m+1)!$

We can rewrite $(r-m+1)!$ as $(r-m+1) \times (r-m)!$ .

Total permutations $= m! \times \frac{(n-m)!}{(r-m)!(n-r)!} \times (r-m+1) \times (r-m)!$

The $(r-m)!$ terms cancel out:

Total permutations $= m! \times \frac{(n-m)!}{(n-r)!} \times (r-m+1)$

Recognize that $\frac{(n-m)!}{(n-r)!}$ is the formula for ${}^{n-m}P_{r-m}$ .

Total permutations $= m! \times {}^{n-m}P_{r-m} \times (r-m+1)$ …

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Q.In the permutations of nnn things, rrr taken together, the number of permutations in which mmm particular things occur together is n−mPr−m×rPm{}^{n-m}P_{r-m} \times {}^{r}P_{m}n−mPr−m​×rPm​.

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Q.In the permutations of $n$ things, $r$ taken together, the number of permutations in which $m$ particular things occur together is ${}^{n-m}P_{r-m} \times {}^{r}P_{m}$ .

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