nPr Formula

nPr formula offers a powerful tool for calculating permutations—the arrangements of distinct elements in a specific order. In this article, we delve into the nPr formula’s significance, properties, mathematical derivation, and diverse applications across mathematics and real-world scenarios. With a focus on order and arrangement, the formula finds utility in fields ranging from probability theory to genetics, demonstrating its versatility and practical relevance.

What is ⁿP_r Formula?

A permutation is an arrangement of all or part of a set of objects, about the order of the arrangement. The nPr formula is used to calculate the number of permutations of n distinct objects taken r at a time. It is denoted mathematically as:

ⁿP_r = n! / (r!(n – r)!)

Where,

• “n” is the total number of items in the set,
• “r” is the number of items to be chosen, and
• “!” denotes factorial, which is the product of all positive integers from 1 to the given number.

nPr Formula

nPr formula is used when we have to choose “r” options out of “n” choices. And the nPr formula is,

P (n, r) = ⁿP_r = _nP_r = n! / (n – r)!

Where,

• n is Total Number of Things
• r is Number of Things that have to be Selected and Arranged

Properties of nPr Formula

Some of the common properties of the ⁿP_r Formula are:

• ⁿP_n = ⁿP_n-1

Proof:

LHS: ⁿP_n = n!/(n-n)! = n!/0! = n!

RHS: ⁿP_n-1 = n!/[n-(n-1)]! = n!/1! = n!

Thus, LHS = RHS

• ⁿP_r = n × ^n-1P_r-1

Proof:

LHS: ⁿP_r = n!/(n-r)!

RHS: n × ^n-1P_r-1 = n × (n-1)!/[(n-1)-(r-1)]! = [n × (n-1)!]/(n-r)! = n!/(n-r)!

Thus, LHS = RHS

• ⁿP_r = ^n-1P_r + ^{r × (n-1)}P_r-1

Proof:

LHS: ⁿP_r = n!/(n-r)!

^n-1P_r = (n-1)!/(n-r-1)!

^{r × (n-1)}P_r-1= r × (n-1)!/[(n-1)-(r-1)]! = r × (n-1)!/(n-r)!

RHS: ^n-1P_r + ^{r × (n-1)}P_r-1 = (n-1)!/(n-r-1)! + r × (n-1)!/(n-r)! = (n-1)!(n-r)/(n-r)! + r × (n-1)!/(n-r)!

⇒^n-1P_r + ^{r × (n-1)}P_r-1= (n-1)!(n-r+r)/(n-r)! = (n-1)!n/(n-r)! = n!/(n-r)!

Thus, LHS = RHS

Derivation of ⁿP_r Formula

Let the n different objects be a₁, a₂, a₃, . . . , a_n.

First place can be filled up by any one of the n objects in n ways.

Second place can be filled up by any one of the remaining (n-1) objects in (n-1) ways.

Third place can be filled up by any one of the remaining (n-2) objects in (n-2) ways.

This continues and goes on till the r^th place is filled.

The number of ways of filling up the rth place = n-r+1

Using Fundamental Counting Principle,

Total permutations (nPr) = n × (n – 1) × (n – 2) × … × (n – r + 1) = (n!)/(n-r)!

This formula calculates the number of ways to arrange ‘r’ elements out of ‘n’ distinct elements without repetition.

ⁿP_r and ⁿC_r Formula

Combination means selection. Here, the order does not matter. Whereas permutation of n different objects taken r at a time = (number of ways of selecting r objects from n different objects) *× (number of ways of arranging the selected r objects) i.e., permutation is the way to arrange some objects.

Formula	Interpretation	Expression
nPr	Permutation of ‘n’ different objects taken ‘r’ at a time	n!/(n-r)!
nCr	Combination of n different objects taken r at a time	n!/(n-r)!(r!)

Where,

• n is the total number of objects,
• r is the number of objects taken at a time, and
• Factorial i.e., n! is the product of all positive integers from 1 to “n.”

Relationship between ⁿP_r and ⁿC_r Formula

As ⁿP_r = (number of ways of selecting r objects from n different objects) × r!

⇒ (number of ways of selecting r objects from n different objects) = ⁿP_r /r!

⇒ ⁿC_r = ⁿP_r /r! = n!/(n-r)!(r!)

Thus, the relationship between ⁿP_r and ⁿC_r Formula is: ⁿC_r = ⁿP_r /r!

Applications of the ⁿP_r formula

Key applications of ⁿP_r Formula are:

• Combinatorial Analysis
  • This is about counting different ways things can be arranged.
  • For example, arranging people in a line, picking a group, or creating passwords with unique characters.
• Probability and Statistics
  • In chance and data analysis, we use arrangements to figure out how likely certain outcomes are.
  • This helps in understanding and predicting probabilities, which is important in statistics.
• Generating Arrangements
  • In computers and encryption, arrangements are used to create unique sequences.
  • This is handy for securing information, shuffling data, or making things random.
• Genetics and Biology
  • In genetics, arrangements are studied to understand gene sequences and variations.
  • This is crucial for genetic research and understanding how living things are put together at a molecular level.
• Game Theory
  • In games and puzzles, arrangements are important for figuring out all the possible moves or solutions.
  • This helps in planning strategies and solving problems when playing board games or puzzles.

Read More,

• Permutation
• Combinations
• Permutation and Combination

Solved Examples on ⁿP_r Formula

Example 1: Suppose you have a deck of 52 playing cards, and you want to find the number of ways to choose 5 cards in a specific order from the deck (i.e., permutations of 5 cards out of 52).

Solution:

Here, n = 52 (total number of cards in the deck) and r = 5 (number of cards to be chosen).

⁵²P₅ = 52!/(52-5)!

⇒ ⁵²P₅ = 52!/(47)!

⇒ ⁵²P₅ = (52 × 51 × 50 × 49 × 48 × 47!)/47!

⇒ ⁵²P₅ = 311,875,200

Therefore, there are 311,875,200 different ways to choose 5 cards in a specific order from a standard deck of 52 playing cards.

Example 2: Seven athletes are participating in a race. In how many ways can the first 3 athletes win the prize?

Solution:

Here, n = 7 (total number of athletes participating in the race) and r = 3 (number of athletes to be chosen to win the prize).

⁷P₃ = 7!/(7-3)!

⇒ ⁷P₃ = 7!/(4)!

⇒ ⁷P₃ = 7 × 6 × 5 × 4!/4!

⇒ ⁷P₃ = 7 × 6 × 5 = 210

Hence, there are 210 ways 3 athletes can win the prize.

Example 3: In how many ways can 6 persons stand in a queue?

Solution:

Here, n = 6 (total number of persons) and r = 3 (number of persons to stand in the queue).

⁶P₆ = 6!/(6-6)!

⇒ ⁶P₆ = 6!/(0)! = 6!/1 = 6! = 6 × 5 × 4 × 3 × 2 × 1 = 720

Hence, there are 720 ways 6 persons can stand in a queue.

Example 4: A student has 12 different books on a shelf. In how many ways can the student arrange 7 books on the top shelf?

Solution:

Here, n = 12 (total number of books) and r = 7 (number of books to be arranged).

¹²P₅ = 12!/(12-5)!

⇒ ¹²P₅ = 12!/(7)!

⇒ ¹²P₅ = 12 × 11 × 10 × 9 × 8 × 7!/7!

⇒ ¹²P₅ = 12 × 11 × 10 × 9 × 8 = 95040

Hence, there are 95040 ways to arrange 7 books on a shelf out of 12 different books.

Practice Problems on ⁿP_r Formula

Problem 1: If ²ⁿ⁺¹P_n-1:^2n-1P_n = 3:5, then find the value of n.

Problem 2: How many different signals can be given using any number of flags from 5 flags of different colors?

Problem 3: Find the sum of all the numbers that can be formed with the digits 2, 3, 4, and 5 taken all at a time.

Problem 4: If ⁿP₅ = 20×ⁿP₃, find the value of n.

Problem 5: How many numbers can be formed from the digits 1, 2, 3, and 4 when repetition is not allowed?

Problem 6: How many 4-letter words, with or without meaning can be formed using the letters in the word LOGARITHMS, if repetition of letters is not allowed?

ⁿP_r Formula: FAQs

1. What are Permutations?

Permutations are arrangements of objects in a specific order. In mathematics, particularly in combinatorics, a permutation of a set is an ordered arrangement of its distinct elements.

2. What is ⁿP_r Formula in Probability?

The ⁿP_r formula is:

ⁿP_r = n! / (n – r)!

Where n is the total number of items and r is the number of items chosen.

3. What is the Value of ⁿP₀?

ⁿP₀ is defined as 1. In combinatorics, the number of ways to arrange 0 elements (which essentially means having an empty set) is considered to be 1.

4. What is the Meaning of r in ⁿP_r?

In ⁿP_r, r is the number of objects chosen and arranged from the total n objects.

5. How do you know when to use ⁿC_r or ⁿP_r?

We use ⁿP_r when order of arrangement matters, whereas we use ⁿC_r when it doesn’t.

6. Is ⁿP_r always greater than ⁿC_r?

No, ⁿP_r is not always greater than ⁿC_r for same value of r and n.