Permutations and Combinations: The different arrangements of objects taking some or all of them at a time is calculated by permutations and combinations. A permutation of a set is an arrangement of its elements into a sequence or a linear order, or if the set is already ordered, a rearrangement of its elements.
A combination is a selection of objects from a collection, such that the order of selection does not matter, or the different groups or selections of several things taking some or all of them at a time are called combinations. In this article, we will learn everything about permutation and combination, important formulas, the difference between the two, and their different types.
Table of Contents
 Permutation and Combination Definitions
 Factorial Notation
 Permutation and Combination Formulas
 Restricted Permutation
 Circular Permutations
 Combinations
 Important Results in Combinations
 Difference Between Permutation and Combination
 Solved Examples – Permutation and Combination
 Summary
 Frequently Asked Questions (FAQ) – Permutations and Combinations
Permutation and Combination Definitions
Permutation: The rearranging of the elements in an ordered set is called the process of permutation. The word “permutation” also refers to the act or process of replacing the linear order of an ordered set.
Combination: The process of selecting items from a collection where the order of selection does not matter is known as a combination, or the different groups or selections of several things taking some or all of them at a time are called combinations.
Factorial Notation
Before going to the different formulas of permutation and combination, one should know about the factorial notation.
The continued product of first \(n\) natural numbers is called the factorial of \(n.\) It is denoted by \(n!\) and is defined as
\(n! = 1.2.3 \ldots \ldots (n – 2)(n – 1)n\)\( = n(n – 1)(n – 2) \ldots \ldots 3.2.1\)
\( = n \times (n – 1)!\)
\( = n \times (n – 1) \times (n – 2)! \ldots \ldots ..\)
Permutation and Combination Formulas
The number of permutations of \(n\) different things taken \(r\) at a time is denoted as \(^n{P_r}\) or \(P(n, r)\) and is defined as \(^n{P_r} = \frac{{n!}}{{(n – r)!}}\), where \(r≤n, r∈W, n∈N.\)
The number of combinations of \(n\) different things taken \(r\) at a time is denoted by \({}^n{C_r}\) and is defined as \(^n{C_r} = \frac{{n!}}{{r! \times (n – r)!}},\) where \(r≤n, r∈W, n∈N\)
Permutation with Repetition
The number of permutations of \(n\) different things taken \(r\) at a time, when repetition is allowed is \({n^r}\).
Permutation of Similar Objects
The number of permutations of \(n\) things taken all at a time, where \(p\) are alike of one kind, \(q\) are alike of the second kind, \(r\) are alike of the third kind, and the rest are different is given by
\(\frac{{n!}}{{p! \times q! \times r!}}\)
Restricted Permutation
Together: The number of permutations of \(n\) different things taken all at a time when \(m\) specified things always come together is given as \((n –m+1)!×m!.\)
NotTogether: The number of permutations of \(n\) different things taken all at a time when \(m\) specified things never come together is given as \(n! –(n –m+1)!×m!.\)
The Rank of Word in a Dictionary
According to the English dictionary, the rank of a word is the position of that word when we arrange them in alphabetic order.
To understand the concept of the rank of a word, we should remember the following steps.
Consider the three letters \(L, M,\) and \(N.\) The number of arrangement of the letters \(L, M,\) and \(N\) is \(3!\) and their order is
\(LMN—1st\) word
\(LNM—2nd\) word
\(MLN—3rd\) word
\(MNL—4th\) word
\(NLM—5th\) word
\(NML—6th\) word
If we ask you, what is the rank of the word \(NLM\)? The answer is \(5th\).
Gap Method
The number of permutations of \(n\) different things taken all at a time when \(m\) specified things are to be arranged such that no two of which are to occur together is \(^{n + 1}{P_m} \times n!\)
Circular Permutations
The permutations of objects in a row are called linear permutations of linear arrangements. If \(n\) different things can be arranged in a row, the linear arrangement is \(n!\) whereas every linear arrangement has a beginning and end, but in circular permutations, there is neither beginning nor end. When clockwise and anticlockwise orders are taken as different, the number of circular permutations of \(n\) different things taken all at a time is \((n – 1)!\)
But, when the clockwise and anticlockwise orders are not different, i.e. the arrangements of beads in a necklace, arrangements of flowers in a garland, etc., the number of circular permutations of \(n\) different things \( = \frac{1}{2} \times \left( {n – 1} \right)!\)
Restricted Circular Permutations
1. If clockwise and anticlockwise arrangements are taken as different, the number of circular permutations of \(n\) other things, taken \(r\) at a time is given by \(\frac{{^n{P_{{\rm{r }}}}}}{r}\)
2. If clockwise and anticlockwise arrangements are not taken as different, the number of circular permutations of \(n\) other things, taken \(r\) at a time is \(\frac{{^n{P_{{\rm{r }}}}}}{{2r}}\)
Combinations
The different groups or selections of several things taking some or all of them are called combinations. The number of combinations of \(n\) other things taken \(r\) at a time is denoted by \(^n{C_r}\) and is defined as \(^n{C_r} = \frac{{n!}}{{r! \times (n – r)!}},\) where \(r≤n, r∈W, n∈N\)
Important Results in Combinations
1. \(^n{C_r}{ = ^n}{C_{n – r}}\)
2. \(^n{C_x}{ = ^n}{C_y} \Rightarrow x = y\) or \(x + y = n\)
3. \(^n{C_r}{ + ^n}{C_{r – 1}}{ = ^{n + 1}}{C_r}\)
4. \(\frac{{^n{C_r}}}{{^n{C_{r – 1}}}} = \frac{{n – r + 1}}{r}\)
5. \(^n{C_0}{ + ^n}{C_1}{ + ^n}{C_2} + \ldots \ldots \ldots { + ^n}{C_n} = {2^n}\)
6. \(^nC_0^2{ + ^n}C_1^2{ + ^n}C_2^2 + \ldots \ldots ..{ + ^n}C_n^2{ = ^{2n}}{C_n}\)
Difference Between Permutation and Combination
We can understand the difference between permutation and combination through the following points.
1. Permutation refers to the arrangement, and combination refers to selection.
2. In permutations, order/sequence of arrangement is considered, unlike in combinations.
3. The permutation of two things from three given things \(p, q, r\) is \(pq, qp, qr, rp, pr, rp.\) The combination of two things from three given things \(p, q, r\) is \(pq, qr, rp\)
4. For permutation of \(r\) objects out of n different objects is given by \(^n{P_r} = \frac{{n!}}{{(n – r)!}}\). For the selection of \(r\) objects out of n different objects is given by \(^n{C_r} = \frac{{n!}}{{r! \times (n – r)!}}\)
Solved Examples – Permutation and Combination
Q.1. Find \(x,\) if \(\frac{x}{{5!}} + \frac{x}{{6!}} = \frac{1}{{7!}}\)
Ans: Given \(\frac{x}{{5!}} + \frac{x}{{6!}} = \frac{1}{{7!}}\)
\( \Rightarrow \frac{x}{{5!}} + \frac{x}{{6 \times 5!}} = \frac{1}{{7 \times 6 \times 5!}}\)
\( \Rightarrow x + \frac{x}{6} = \frac{1}{{7 \times 6}}\)
\( \Rightarrow \frac{{7x}}{6} = \frac{1}{{7 \times 6}}\)
\( \Rightarrow x = \frac{1}{{49}}\)
Q.2. How many \(3\) digit even numbers can be formed by using the digits \(1, 2, 3, 4,\) and \(5\)?
Ans: The given digits are \(1, 2, 3, 4,\) and \(5\). A number will be even if the unit digit is even.
Thus, the unit digit can be filled in \(2\) different ways.
Hence, the even total numbers \( = 2{ \times ^4}{P_2}\)
\( = 2 \times \frac{{4!}}{{2!}}\)
\( = 2 \times \frac{{4 \times 3 \times 2!}}{{2!}}\)
\( = 2 \times 4 \times 3\)
\(=24\)
Therefore, \(24\) three digits even number can be formed using the given digits
Q.3. How many \(4\) digit numbers can be formed by using the digits \(1\) to \(9,\) if repetition of digits is not allowed ?
Ans: Here, order matters; for example, \(1234\) and \(1324\) are two different numbers. Therefore, there will be as many \(4\) digit numbers as there are permutations of \(9\) different digits taken \(4\) at a time.
Therefore, the required \(4\) digit numbers \({ = ^9}{P_4} = \frac{{9!}}{{(9 – 4)!}} = \frac{{9!}}{{5!}} = \frac{{9 \times 8 \times 7 \times 6 \times 5!}}{{5!}} = 3024\)
Q.4. Find the number of different \(8\) letter arrangements that can be made from the letters of the word DAUGHTER so that all vowels occur together
Ans: There are \(8\) different letters in the word DAUGHTER, in which there are \(3\) vowels, namely, \(A, U,\) and \(E\). Since the vowels have to occur together, we can, for the time being, assume them as a single object \((AUE)\). This single object, together with 5 remaining letters (objects) will be counted as \(6\) objects. Then we count permutations of these \(6\) objects taken all at a time. This number would be \(^6{P_6} = 6!\). Corresponding to each of these permutations, we shall have \(3!\) permutations of the three vowels \(A, U, E\) took all at a time. Hence, by the multiplication principle, the required number of permutations \(=6!×3!=4320\)
Q.5. A committee of \(3\) persons is to be constituted from a group of \(2\) men and \(3\) women. In how many ways can this be done? How many of these committees would consist of \(1\) man and \(2\) women?
Ans: Here, the order does not matter. Therefore, we need to count combinations. There will be as many committees as there are combinations of \(5\) different persons taken \(3\) at a time. Hence, the required number of ways \({ = ^5}{C_3} = \frac{{51}}{{3!2!}} = \frac{{4 \times 5}}{2} = 10\)
Now, \(1\) men can be selected from \(2\) men in \(^2{C_1}\) ways and \(2\) women can be selected from \(3\) women in \(^3{C_2}\) ways. Therefore, the required number of committees \({ = ^2}{C_1}{ \times ^3}{C_2}\)
\( = \frac{{21}}{{1!1!}} \times \frac{{3!}}{{2!1!}} = 6\)
Summary
In this article, we studied factorial notation, definitions of permutation, and combination, the difference between permutations and combinations and some solved examples.
Frequently Asked Questions (FAQ) – Permutations and Combinations
Q.1. How do you calculate permutations and combinations?
Ans: We will calculate permutations and combinations using the below formulas.
Permutation: \(n{P_r} = \frac{{n!}}{{(n – r)!}}\)
Combination: \(^n{C_r} = \frac{{n!}}{{r! \times (n – r)!}}\)
Q.2. What is the main difference between permutation and combination?
Ans: Permutations are used when an arrangement is required. Combinations are used when selection is required.
Q.3. How do you solve permutations?
Ans: Permutation is solved using the formula \(^n{P_r} = \frac{{n!}}{{(n – r)!}}\)
Q.4. What is \(r\) in the combination formula?
Ans: In the combination formula \(^n{C_r} = \frac{{n!}}{{r! \times (n – r)!}};n\) represents the number of items, and \(r\) represents the number of items being chosen at a time.
Q.5. What is \(r\) in the permutation formula?
Ans: In the permutation formula \(^n{P_r} = \frac{{n!}}{{(n – r)!}};n\) is the total items in the set, and \(r\) is the items taken for the permutation.
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