If ${}^{n}P_{r} = 840$, ${}^{n}C_{r} = 35$, then $r =$ ______.

> [!TLDR] > The fundamental relationship between permutations and combinations, ${}^{n}P_{r} = {}^{n}C_{r} \cdot r!$, allows us to find $r!$ by dividing the given permutation value by the combination value. Solving $r! = 24$ yields $\boxed{r=4}$. When we talk about permutations and combinations, we are dealing with ways to select and arrange items from a larger set. The key difference lies in whether the order of selection matters. * **Permutations (${}^{n}P_{r}$):** This represents the number of ways to *arrange* $r$ distinct items chosen from a set of $n$ distinct items. Here, the order of the items is important. For example, selecting A then B is different from selecting B then A. * **Combinations (${}^{n}C_{r}$):** This represents the number of ways to *select* $r$ distinct items from a set of $n$ distinct items. Here, the order of the items does not matter. Selecting A then B is considered the same as selecting B then A. The relationship between these two concepts is crucial. If we first *select* $r$ items from $n$ items (which can be done in ${}^{n}C_{r}$ ways), and then *arrange* those $r$ selected items (which can be done in $r!$ ways), we get the total number of permutations of $r$ items from $n$. > [!FORMULA] > The relationship between permutations and combinations is given by: > ${}^{n}P_{r} = {}^{n}C_{r} \cdot r!$ This formula makes intuitive sense: for every unique group of $r$ items you can choose (a combination), there are $r!$ ways to arrange those specific $r$ items. Multiplying these two gives the total number of ordered arrangements (permutations). Let's use this understanding to solve the problem. 1. **Identify the given values:** We are given the value of the permutation: ${}^{n}P_{r} = 840$. We are also given the value of the combination: ${}^{n}C_{r} = 35$. …

If ⁿPr = 840, ⁿCr = 35, then r = ___.

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{}^{n}P_{r} = 840

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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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The fundamental relationship between permutations and combinations, ${}^{n}P_{r} = {}^{n}C_{r} \cdot r!$ , allows us to find $r!$ by dividing the given permutation value by the combination value. Solving $r! = 24$ yields $\boxed{r=4}$ .

When we talk about permutations and combinations, we are dealing with ways to select and arrange items from a larger set. The key difference lies in whether the order of selection matters.

Permutations ( ${}^{n}P_{r}$ ): This represents the number of ways to arrange $r$ distinct items chosen from a set of $n$ distinct items. Here, the order of the items is important. For example, selecting A then B is different from selecting B then A.
Combinations ( ${}^{n}C_{r}$ ): This represents the number of ways to select $r$ distinct items from a set of $n$ distinct items. Here, the order of the items does not matter. Selecting A then B is considered the same as selecting B then A.

The relationship between these two concepts is crucial. If we first select $r$ items from $n$ items (which can be done in ${}^{n}C_{r}$ ways), and then arrange those $r$ selected items (which can be done in $r!$ ways), we get the total number of permutations of $r$ items from $n$ .

{}^{n}P_{r} = {}^{n}C_{r} \cdot r!

This formula makes intuitive sense: for every unique group of $r$ items you can choose (a combination), there are $r!$ ways to arrange those specific $r$ items. Multiplying these two gives the total number of ordered arrangements (permutations).

Let's use this understanding to solve the problem.

Identify the given values: We are given the value of the permutation: ${}^{n}P_{r} = 840$ . We are also given the value of the combination: ${}^{n}C_{r} = 35$ . …

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Q.If nPr=840{}^{n}P_{r} = 840nPr​=840, nCr=35{}^{n}C_{r} = 35nCr​=35, then r=r =r= ______.

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Q.If ${}^{n}P_{r} = 840$ , ${}^{n}C_{r} = 35$ , then $r =$ ______.

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