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Permutation Formula is nPr = (n!) / (n – r)!. It is used to determine the various numbers of arrangements that can be created by selecting r items from the total of n items. Permutations are helpful in forming different words, numerical arrangements, seating arrangements, and for any other circumstances involving different arrangements.
Key Terms: Permutation, Factorial, Combination, Principle of Counting, Series
Permutation Formula
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The formula of permutation is nPr = (n!) / (n – r)!. Permutations represent a variety of arrangements that can be possible in a group. In the case of permutations, order is absolutely essential. This sets it apart from a combination, which is a concept where order doesn't matter. To some degree, permutations are a form of ordered combinations.
According to the permutation formula, the permutation of "r" items picked from "n" objects is equal to the factorial of n divided by the factorial of difference of n and r.
\(P (n,r) = \frac{n!}{(n-r)!}\)
where
- n = total items in the set;
- r = items taken for the permutation;
- "!" denotes taking the factorial.
Permutations and Combinations Detailed Video Explanation
Example of PermutationQues: How many different ways can four boys from a class of 15 be made stand for an assembly? Answer: There are P (15,4) possible permutations of 4 students from a group of 15. P (15,4) = 15! ÷ (15 - 4)! = 15! ÷ 11! = 15×14×13×12×11! ÷ 11! = 15×14×13×12 = 32760 different line-ups. |
What is Factorial?
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The factorial function ('!') is used in the calculation of permutations and combinations, it just means to multiply a series of descending natural numbers.
It means all the numbers in the series are multiplied together (i.e., from 1 to n).
Example of Factorial5! =5× 4 × 3 × 2 × 1 = 120 8! = 8×7 × 6 × 5 × 4 × 3 × 2 × 1 = 40,320 1! = 1 0! = 1 Factorial of n natural numbers. n! = 1 × 2 × 3 × 4 × .......× n |
Types of Permutations
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There are many distinct types of permutations. The following are the two main categories of permutations:
When repetition is not Allowed
P is a permutation or arrangement of r things from a set of n things without replacement.
Formula
The permutations formula used when 'r' things from 'n' things have to be arranged without repetitions is nothing but the nPr or P (n, r) formula given as:
P (n, r) = (n!) / (n - r)!
When repetition is Allowed
P is a permutation or arrangement of r things from a set of n things when duplication is allowed.
Formula
The permutations formula used when 'r' things from 'n' things have to be arranged with repetitions is just nʳ. This is because each of the 'r' things can be selected in 'n' different ways, given as:
n ×n ×n× .... ×n (r times) = nʳ
What are Combinations?
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According to its definition, a combination is "An arrangement of objects where the order in which the objects are selected is irrelevant." The phrase "Selection of things" denotes a situation in which the chronological order of events is not important.
The term n Cr is the number of combinations of n different items taken r at a time, and it is given as,
\(n_{c_r} = \frac{n!}{r!(n-r)!}\)
where,
- C (n, r) = No. Of combinations
- n = number of objects total in the set r
- r = number of choosing objects from the set and 0 ≤ r ≤ n.
This forms the general combination formula which is ⁿCr formula.
Solved ExampleExample: Out of a group of 5 people, a pair needs to be formed. The formula below can be used to get the total number of possible combinations. C (5,2) = 5! / 2! (5-2)! = 5! / 2! × 3! =120/ 2×6 = 10 |
Difference between Permutations and Combinations
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The list of differences between Permutations and Combinations are:
| Permutations | Combinations |
|---|---|
| A selection of r objects from a set of n objects in which the order of the selection matters. | The variety of combinations that can be made using r objects out of a total of n objects in which the order of selection doesn’t matters. |
| Data is chosen from a list | Data is chosen from a group |
| There is an arrangement of data | There is a selection of data |
| They are defined as ordered elements as order matters in Permutation. | They are defined as unordered sets as order doesn't matter in Combination. |
| Multiple permutations are possible from one combination | One combination is possible from one permutation |
Principle of Counting
The basic counting principle is a set of rules for counting all the possibilities in which an event may occur or the total possibilities of outcomes. It implies that if there are n ways of doing a thing and m ways to do another, then their product can be used to find the total number of ways to do both the tasks. It is represented as n × m.
Solved ExampleExample: A classroom has six students and six benches for them to sit on. Every month, the class teacher makes them sit in a different spot. How many different ways can she find to make them sit in class? Solution: Given that, 6 children are present, and there are 6 benches on which they can sit. By using the principle of counting, they can thus be made to sit in as many different ways as possible by their teacher as - 6 × 6 = 36 ∴ There are 36 ways. |
Things to Remember
- Formula 1: Factorial of n natural numbers.
n! = 1 × 2 × 3 × 4 × ...× n
- Formula 2: Permutation Formula for r things taken from n things.
nPr = (n!) / (n - r)!
- Formula 3: The relationship between permutations and combinations for r things taken from n things.
nPr = r! × n Cr
- A permutation is the number of ways a set can be arranged or the number of ways things can be arranged.
- With a permutation, the order of numbers matters.
- The main types of permutations are those with repetition and those without, although other less common types include permutations with multi-sets and circular permutations.
- It is possible to have multiple permutations from a single combination.
- Permutations are distinct from combinations, which is a selection of data from a group where order doesn't matter.
Previous Year Questions
- How many four digit numbers abcd exist such that a is odd, b is divisible by 3, c is even…? [KEAM]
- Let Tn be the number of all possible triangles formed by joining vertices of an n-sided regular…? [JEE 2013]
- Let A and B be two sets containing four and two elements respectively. Then the number of subsets…? [JEE 2015]
- If nCr−1=28, nCr=56 and nCr+1=70, then the value of r is equal to…? [KEAM]
- The number of words that can be formed by using all the letters of the word PROBLEM only one is…? [KEAM]
- If all the words (with or without meaning) having five letters, formed using the letters…? [JEE 2016]
- If a, b and c are the greatest values of 19Cp, 20Cq and 21Cr respectively, then…? [JEE 2020]
- Consider a class of 5 girls and 7 boys. The number of different teams consisting of 2 girls and…? [JEE 2019]
- All possible numbers are formed using the digits 1,1,2,2,2,2,3,4,4 taken all at a time…? [JEE 2019]
- An eight digit number divisible by 9 is to be formed using digits from 0 to 9 without repeating…? [JEE 2014]
Sample Questions
Ques. If any box can hold any number of balls, with the exception of ball 3, which can only be placed in box 3 or box 4, how many different ways may 7 distinct balls be distributed among 5 different boxes? (3 Marks)
Ans. 1st ball can be placed in any of the 5 boxes.
2nd ball can be placed in any of the 5 boxes.
Ball 3 can either be placed into box 3 or box 4.
Hence, 3rd ball can be placed into any of these 2 boxes.
We can place the 4th in any of the 5 boxes.
We can place the 5th ball in any of the 5 boxes.
We can place the 6th in any of the 5 boxes.
We can place the 7th ball in any of the 5 boxes.
Hence, required number of ways
=5×5×2×5×5×5×5
=2×5⁶
Ques. How many different combinations do you get if you have 4 items and choose 2? (2 Marks)
Ans. In this instance, "n" denotes the total number of items in the collection, and "r" denotes the number of things that were selected then,
C (n, r) = n! / r! (n – r)!
= 4! / 2! (4 – 2)!
= 4! /2! * 2!
= 4 x 3 x 2 x 1 / 2 x 1 * 2 x 1
= 24 / 4
= 6
The solution is 6.
List of possible combinations:
{1, 2}
{1, 3}
{1, 4}
{2, 3}
{2, 4}
{3, 4}
Ques. How many words with three consonants and two vowels can be created from seven consonants and four vowels? (5 Marks)
Ans. The number of possible ways to choose 3 consonants from 7 is
= 7C3
The number of possible ways to choose 2 vowels from 4
= 4C2
= Number of ways to choose 2 vowels from 4 and 3 consonants from
= 7C3 × 4C2
= [(7×6×5) / (3×2×1)] x [(4×3) / (2×1)]
= 210
= It implies that we can have 210 groups, each of which has a total of 5 letters (3 consonants and 2 vowels).
Number of possible combinations for grouping five letters together
= 5! = 5×4×3×2×1=120=5! = 5×4×3×2×1=120
Hence, required number of ways
=210×120=25200
Ques. The vowels in the word "OPTICAL" can be arranged in how many distinct manners that they always come together? (3 Marks)
Ans. The word 'OPTICAL' consists of 7 letters. It contains the vowels "O," "I," and "A," which need to all come together. All such 3 vowels can therefore be combined and regarded as a single character that is PTCL (OIA).
We can therefore assume that there are 5 total letters, each of which is unique. That is, PTCL(OIA).
Hence we assume the total letters as 5 and all these letters are distinct.
So, the variety of ways these letters can be arranged
=5!=5×4×3×2×1=120=5!=5×4×3×2×1=120
All the 3 vowels (OIA) are different
Number of ways to arrange these vowels among themselves
=3!=3×2×1=6=3!=3×2×1=6
Hence, required number of ways
=120×6=720
Ques. A school's working day consists of 6 sessions. How many different ways are there to arrange five different subjects so that each receives at least 1 period? (3 Marks)
Ans. 5 subjects can be arranged in 6 periods in 6P5 ways.
Any of the 5 subjects can be organized in the remaining period (5C1 ways).
Two subjects are alike in each of the arrangement. So, we dividing it by 2! to avoid overcounting.
Total number of arrangements are
(6P5× 5C1)/ 2! = 1800
Ques. How many 6-digit telephone numbers can be formed if each number starts with 35 and no digit appears more than once? (2 Marks)
Ans. The first two places can only be filled by 3 and 5 respectively and there is only 1 way for doing this.
Considering that no digit occurs more than once. Hence, we have 8 digits remaining (0,1,2,4,6,7,8,9)
So, the next 4 places can be filled with the remaining 8 digits in 8P4 ways.
Total number of ways = 8P4 =8×7×6×5=1680
Ques. 10 chair patterns and 8 table patterns are available to an organiser. How many different ways can he create a set with a table and chair? (2 Marks)
Ans. He has 10 patterns of chairs and 8 patterns of tables
A chair can be selected in 10 ways.
A table can be selected in 8 ways.
Hence one chair and one table can be selected in 10×810×8 ways =80
=80 ways
Ques. There are 25 buses operating between P and Q. How many distinct bus routes are there to travel from P to Q and back? (3 Marks)
Ans. He can go in any of the 25 buses (25 ways).
Since he cannot come back in the same bus, he can return in 24 ways.
Total number of ways =25×24=600
Ques. There are 5 yellow, 4 green and 3 black balls in a bag. All the 12 balls are drawn one by one and arranged in a row. Find out the variety of combinations that are feasible, if any. (3 Marks)
Ans. Number of different arrangements possible
\(= \frac{12!}{5! 4! 3!}\)
\(= \frac{12×11×10×9×8×7×6×5×4×3×2}{(5×4×3×2)(4×3×2)(3×2)}\)
\(= \frac{12×11×10×9×8×7×6}{(4×3×2)(3×2)}\)
\(= \frac{11×10×9×8×7}{2}\)
=11×10×9×4×7
=252×11×10
=27720
Ques. There are 12 black , 7 red, and 6 blue balls in a box. How many methods are there to pick one or more balls? (2 Marks)
Ans. Number of ways in which one or more objects can be selected out of S1 alike objects of one kind, S2 alike objects of second kind and S3 alike objects of third kind
= (S1 + 1) (S2 + 1) (S3 + 1) – 1
Hence, require number of ways
= (12+1) (7+1) (6+1) −1= (13×8×7) −1=728−1
=727
Ques. Five balls need to be placed in three boxes. Each box can hold all the five balls. In how many ways can the balls be placed in the boxes if no box can be empty, all balls are identical but all boxes are different? (3 Marks)
Ans. Here n = 3, k = 5.
Hence, as per the above formula, required number of ways
= (k-1) C(n-1)
= 4C2
= 6
Ques. If every boy must receive at least one toy, how many different ways may 30 matching toys be distributed among 10 boys? (3 Marks)
Ans. Toys are the same here, but boys are different.
If each boy must receive at least one toy, how many different ways may 30 identical toys be distributed among 10 boys?
= Number of ways in which 30 identical balls can be distributed into 10 boxes if each box must contain at least one ball
Hence, this problem can be solved using the formula given at the top.
n = 10, k = 30.
Hence, required number of ways
= (k-1)C(n-1) = 29C9
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