Difference between permutations and combinations: The difference between permutation and combination can be understood by knowing the different situations where these concepts are used. We primarily use combinations for selections of distinct objects, and to find the number of possible arrangements of dissimilar or distinct objects, we use permutations.

In simpler words, when there is no specific order, it is called combinations, and when the order matters, it is called permutations. We can also say that a permutation is an ordered combination. In this article, let us learn in detail about the difference between Permutation and Combination.

Relation Between Permutation and Combination

The formula for \(r\) objects taken from \(n\) objects is such that their permutation is equal to the product of \(r\) factorial and combination.

Therefore, \({}^n{P_r} = \frac{{n!}}{{\left( {n – r} \right)!}}\)
\( \Rightarrow {}^n{P_r} = \frac{{r!n!}}{{r!\, \cdot \,\left( {n – r} \right)!}}\)
\(\therefore \,\,{}^n{P_r} = \,r!\,\, \times \,{}^n{{\text{C}}_r}\)

Permutations

The permutation is an arrangement of objects in a definite order. The objects of sets are arranged here in a sequence or linear order.

Example: If there are \(3\) objects \(a,\,b\) and \(c\), then the permutations of these objects, taking \(2\) at a time is \(ab,\,ba,\,bc,cb,\,ac,\) and \(ca\).

So, the number of permutations of three different objects taken \(2\) at a time is \(6\).

If \(n\) and \(r\) are two positive integers, such that \(0 \leqslant r \leqslant n\) then the number of permutations of \(n\) dissimilar or distinct objects taken \(r\) at a time is given by:
\(P\left( {n,r} \right)\) or \({}^n{P_r}\)

Thus, \(P\left( {n,r} \right)\) or \({}^n{P_r} = \) total number of permutations of \(n\) dissimilar objects taken \(r\) at time.
\(\therefore P\left( {n,r} \right)\, = \,{}^n{P_r} = \frac{{n!}}{{\left( {n – r} \right)!}}\)

Combinations

We are familiar with the arrangements of a certain number of objects by taking some of them or all at a time. Most of the time, we are not interested in arranging the objects, but we are more concerned with selecting some objects from a given number of objects. In other words, we do not want to specify the order of selected objects.

For example, a company may want to select \(3\) out of \(10\) applicants, and a student may want to choose \(2\) books from the library at a time.

Suppose we have to select \(3\) objects out of \(A,\,B,\,C\) and \(D\). We may choose \(A,\,B,\,C\) or \(A,B,D\) or \(A,\,C,D\) or \(B,C,D\).

Note that we have not listed \(A,\,B,\,C\,;\,B,C,\,A;\,C,A,B;\,\,B,A,C;C,B,A\) and \(A,C,B\) separately here, because they represent the same selection \(A,\,B,\,C\). But, they give rise to different arrangements.

Hence, we can say that in a selection, the order in which objects are arranged is immaterial.

In general, the number of all combinations of \(n\) objects, taken \(r\) at a time is generally denoted by:
\(C\left( {n,r} \right),\) or \({}^n{{\text{C}}_r}\)

Thus, \({}^n{{\text{C}}_r}\) or \({\text{C}}\,\left( {n,r} \right) = \) number of ways of selecting \(r\) objects from \(n\) objects.

Here, \({}^n{{\text{C}}_r}\) is defined only when \(n\) and \(r\) non-negative integers, such that \(0 \leqslant r \leqslant n\).

Number of all combinations of \(n\) distinct objects, taken \(r\) at a time is given by:
\({}^n{{\text{C}}_r} = \frac{{n!}}{{\left( {n – r} \right)!r!}}\)

Difference Between Permutation and Combination

Permutation	Combination
In permutation, not only a selection is made, but also an arrangement in a definite order is considered.	In combination, only selection is made
In permutation, order is essential.	In combination, the order does not matter.
Permutations are generally used for objects of a different kind	Combinations are generally used for objects of the same kind
Permutation of \(2\) objects from \(3\) given objects \(a,b,c\) is \(ab,ba,bc,cb,ac,ca\).	Combinations of \(2\) objects from \(3\) objects \(a,b,c\) is \(ab,bc,ca\).
For different possible arrangements of \(r\) objects taken from \(n\) objects is \({}^n{P_r} = \frac{{n!}}{{\left( {n – r} \right)!}}\)	For different possible selections of \(r\) objects taken from \(n\) objects is \({}^n{C_r} = \frac{{n!}}{{r!\left( {n – r} \right)!}}\)

Key Differences Between Permutation and Combination

1. Permutation refers to several ways of arranging the set of objects in a particular order. Combination is the several ways of choosing objects from a large pool of objects such that their order is irrelevant,
2. The primary difference is order, placement, and position, i.e., in permutation characteristics mentioned above does matter, which does not matter in combination.
3. Permutation denotes several ways to arrange objects, people, digits, alphabets, colours, etc. On the other hand, the combination indicates different ways of selecting menu objects, food, clothes, subjects, etc.
4. The permutation is nothing but an ordered combination, while the combination implies unordered sets.
5. Many permutations can be obtained from a single combination. Conversely, only a single combination can be obtained from a single permutation.
6. Permutation answers, how many different arrangements can be created from a given set of objects? As opposed to the combination which explains, how many different groups can be picked from a larger group of objects?

Solved Examples

Q.1. How many four-digit numbers are there with distinct digits?
Sol:
The total number of arrangements of ten digits \(0,1,2,3,4,5,6,7,8,9\) taking \(4\) at a time is \({}^{10}{P_4}\) . But, these arrangements also include number \(0\) at thousand’s place. Such numbers are not four-digit numbers. When \(0\) is fixed at thousand’s place, we have to arrange the remaining \(9\) digits by taking \(3\) at a time. The number of such arrangements is \({}^{9}{P_3}\) . So, the total number of numbers having \(0\) at thousand’s place \( = \,{}^9{P_3}\).
Hence, the total number of four-digit numbers \( = \,{}^{10}{P_4} – {}^9{P_3}\)
\( = \frac{{10!}}{{\left( {10 – 4} \right)!}} – \frac{{9!}}{{\left( {9 – 3} \right)!}}\)
\( = \frac{{10!}}{{6!}} – \frac{{9!}}{{6!}}\)
\( = \frac{{10 \times 9 \times 8 \times 7 \times 6!}}{{6!}} – \frac{{9 \times 8 \times 7 \times 6!}}{{6!}}\)
\( = 5040 – 504\)
\( = 4536\)

Q.2. In how many ways \(7\) can pictures be hung from \(5\) picture nails on a wall?
Sol:
The number of ways in which \(7\) pictures can be hung from \(5\) picture nails on a wall is the same as the number of arrangements of \(7\) objects, taking \(5\) at a time.
Hence, the required number \( = \,{}^7{P_5}\)
\( = \frac{{7!}}{{\left( {7 – 5} \right)!}}\,\)
\( = \frac{{7!}}{{2!}}\)
\(=2520\)

Q.3. From a class of \(32\) students, \(4\) are to be chosen for the competition. In how many ways can this be done?
Sol:
Out of \(32\) students, \(4\) students can be selected in \({}^{32}{C_4}\) ways
Therefore, the required number of ways \( = {}^{32}{C_4}\)
\( = \frac{{32!}}{{28!4!}}\)
\( = \frac{{32 \times 31 \times 30 \times 29 \times 28!}}{{28!4!}}\)
\( = \frac{{32 \times 31 \times 30 \times 29}}{{24}}\)
\( = 8 \times 31 \times 5 \times 29\)
\( = 35960\)

Q.4. If there are \(12\) persons in a party, and if each two of them shake hands, how many handshakes happen in the party?
Sol:
It is to note here that it is counted as one handshake when two persons shake hands, not two. So, this is a problem with combinations.
Therefore, the total number of handshakes is the same as the number of ways of selecting \(2\) persons among \(12\) .
\({}^{12}{C_2} = \frac{{12!}}{{10! \times 2!}}\)
\( = \frac{{11 \times 12}}{{1 \times 2}}\)
\(=66\)

Q.5. A question paper has two parts, Part A and Part B, each containing \(10\) questions. If a student has to choose \(8\) from Part A and \(5\) from Part B, in how many ways can he choose the questions?
Sol:
There are \(10\) questions in Part A, out of which \(8\) questions can be chosen in \({}^{10}{C_8}\) ways.
Similarly, \(5\) questions can be chosen from part B containing \(10\) questions in \({}^{10}{C_5}\) ways.
Hence, the required total number of ways of selecting \(8\) questions from part A and \(5\) from part B \( = {}^{10}{C_8} \times {}^{10}{C_5}\)
\( = \frac{{10!}}{{8! \times 2!}}\,\,\, \times \frac{{10!}}{{5! \times 5!}}\)
\(=11340\)

Summary

The permutation is an arrangement of objects in a particular way or order. While dealing with permutation, one should concern about the selection as well as arrangement. In other words, the total number of permutations of \(n\) different objects taken \(r\) at a time is given by \({}^n{P_r} = \frac{{n!}}{{\left( {n – r} \right)!}}\). The combination is a selection of a part of a set of objects or a selection of all objects when the order doesn’t matter. The total number of combinations of \(n\) objects taken \(r\) at a time can be calculated as \({}^n{C_r} = \frac{{n!}}{{\left( {r!} \right)\left( {n – r} \right)!}}\). Then some of key differences between permutations and combinations are discussed.

Frequently Asked Questions (FAQs)

Q.1. What do you mean by permutations and combinations?
Ans: Permutations is an arrangement of objects in a particular order, while combinations is a grouping of objects in which order does not matter.

Q.2. What is the formula for permutations and combinations?
Ans: The number of permutations of \(n\) distinct objects taken \(r\) at a time is given as \({}^n{P_r} = \frac{{n!}}{{\left( {n – r} \right)!}}\). The number of combinations of \(n\) distinct objects taken \(r\) at a time is given as \({}^n{C_r} = \frac{{n!}}{{\left( {r!} \right)\left( {n – r} \right)!}}\)

Q.3. What is the relation between permutation and combination?
Ans: Combinations \( = \) The selection of \(r\) objects out of \(n\) distinct objects
Permutations \( = \) The arrangement of \(r\) objects out of  \(n\) distinct objects
So, \({}^n{C_r} = \frac{{n!}}{{\left( {r!} \right)\left( {n – r} \right)!}}\)  and \({}^n{P_r} = \frac{{n!}}{{\left( {n – r} \right)!}}\).
Hence, arrangements \( = \,r!\, \times \) selections

Q.4. Give examples of permutations and combinations.
Ans: The number of permutations of \(3\) letter words which can be formed by using the letters of the word SPIKE, is \({}^5{P_3} = \frac{{5!}}{{\left( {5 – 3} \right)!}} = 60\)
The number combinations of \(3\) objects taken \(2\) at a time is \({}^3{C_2} = \frac{{3!}}{{2!\, \times 1!}} = 3\)

Q.5. What are areas in Mathematics where permutation and combination are used?
Ans: The concepts of permutations and combinations are mainly used in probability, sets, relations, and functions. The different arrangements can be found with the help of permutations, and the different groups can be found with the help of combinations.

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