CBSE Class 11 Maths Revision Notes Chapter 7

Class 11 Mathematics Revision Notes for Chapter-7 Permutations and Combinations

Extramarks Revision Notes for Class 11 Mathematics Chapter 7 gives a thorough summary of the concepts covered in Chapter Permutations and Combinations. Its ease of access and simple language allow students to maximise exam preparation and boost their confidence before examinations. 

One of the challenging chapters in Mathematics is Permutations and Combinations. Therefore, having reliable notes that can help students navigate such challenging portions of the curriculum is essential.

Revision Notes CBSE Class 11 Mathematics Notes Chapter 7 Permutations and Combinations

What are Permutations and Combinations?

Permutations and combinations define the various ways in which a specific set of data can be structured. This is accomplished by selecting elements from a collection or by forming subsets. According to the definitions, permutation is the process of selecting items or data from a given group. Extramarks Revision Notes for Class 11 Mathematics Chapter 7 contains further information which is concise and simple and is easily accessible for students to understand efficiently.

A permutation is also defined as the phenomenon of reordering data elements that were previously present in order. It can be found in almost every area of mathematics, as seen by looking at the broad field of Mathematics.

Combinations are simply a method for picking elements from a collection. The order of choosing is not thought to have any significance in this procedure. Individuals can also calculate the entire number of possible combinations. It would be helpful if you remembered that this is only true in minor circumstances. Combinations can also be defined as a collection of n items taken k at a time. This should be done once and only once. In circumstances where repetition is permitted, the words k-selection or k-combination with repetition are typically used.

Permutation Formula

Permutation is defined as the selection of r items from a set of n items. This is done without the use of a replacement. In the case of permutation, the order of events is also important. The formula given below can describe all of this.

nPr = (n!) / (n – r)!

Combination Formula

A combination is defined as the selection of r things from a set of n things. All of this is done without replacement in the case of combination, and the order of things is irrelevant. The formula can express all of this information:

nCr = nPr / r! = n! / r! (n – r) !

Difference Between Permutation and Combination

Permutations Combinations
Task of arranging groups like colours, digits, alphabets etc. Selection of subjects like food, clothing, menu items, teams, etc. 
Example- picking a team captain or a picture. Example- picking any three team members randomly.
Deciding on two favourite colours in a particular order on a brochure.  Select any two colours from a particular brochure.
Picking winners for first, second and third places from a team. Picking any three winners randomly for an award from a team.

Fun Facts about Permutation and Combination

Permutations and combinations are useful in both academic and everyday lives. To make our lives easier, we can all employ permutation and combination.

To pick colours to design a room, for instance, or to answer problems in descending order of difficulty. Along with the Extramarks Revision Notes, these apps will help students comprehend concepts more clearly.

FAQs (Frequently Asked Questions)

The formula for calculation permutations is: nPr = n! / (n – r)!

The formula for calculating combinations is: nCr = n! / r!(n−r)!

The fundamental principle of multiplication:

Three distinct events will all take place simultaneously in m×n×p different ways respectively if one of the events happens in m different ways, the second one in n different ways, and the third one in p different ways.

Fundamental principle of addition: 

If two tasks are such that the first task can be completed separately in m different ways and the second task can be completed independently in n different ways, then any of the two tasks can be completed independently in (m+n) different ways.

Division Of Items Into Groups Of Unequal Sizes

A set of distinct objects given by(m+n) can be divided into two unequal groups containing m and n objects.

(m+n)!/m!n!.

A set of distinct objects given by(m+n+p) can be divided into three unequal groups containing m,n and p objects.

m+n+pCm.n+pCm=(m+n+p)!/m!n!p!.

A set of distinct objects given by (m+n+p) can be divided into three persons in the groups containing m,n and p objects are

=(No. of ways to divide)×(No. of groups)!

=(m+n+p)!/m!n!p!×3!.

Divisions Of Objects Into Groups Of Equal Sizes

Without taking into account any order, we can evenly divide mn distinct Objects into m groups, each containing n different objects, in ((mn)!/(n!)m)1/m! ways.

We can equally divide mn different things into m groups, each containing n different objects, while considering the order in  (((mn)!/(n!)m)1/m!)m!=(mn)!/(n!)m ways.

If there are n lines that are not parallel and not concurrent, then nC2 is the number of intersection points. Similar to this, if there are n points, then the number of lines will be nC2 so that no three points are collinear.

If there are m collinear points in n number of points, then the number of straight lines is given by nC2- mC2+1.

If a polygon has n vertices and none of them is collinear, then to calculate the number of diagonals, we can use  nC2-n= n(n-3)/2.

If we have n points and none of them is collinear, the number of triangles that can be made is nC3.

The number of triangles that can be built from n points, where m of them are collinear, is equal to nC3-mC3.

If a polygon has n vertices, the number of triangles that can be formed such that no triangle side is common to that of a polygon side is given by the formula nC3−nC1−(nC1×n−4C1).

The number of parallelograms that can be created from two sets of parallel lines, where one set has n parallel lines and the other set has m parallel lines, is given by nC2×mC2.

The number of squares that can be created from two sets of parallel lines with equal spacing, where one set has n parallel lines and the other set has m parallel lines, is given by ∑m−1r−1(m−r)(n−r);(m<n).

Let’s look at the equation x1+x2+x3+x4+….+xr=n, where x1, x2, x3, x4,…., xr and n are non-negative integers. This equation can be seen as dividing n identical objects into r groups.

In total, there will be n+r−1Cr−1 . non-negative integral solutions to the equation x1+x2+x3+x4+….+xr=n.

The total number of solutions to the problem will be n−1Cr−1 . in the set of natural numbers N.

By adding an artificial or dummy variable xm+1 such that xm+10, we can solve equations of form x1+x2+x3+x4+…..+xm≤n. The equation will be changed to x1+x2+x3+x4+…..+xm+xm+1=n , and the number of solutions can then be determined in the same manner as stated in the points above.

If we have to divide xn identical things among r people so that each person gets 0, 1, or 2 items until it is ≤n, then n+r1Cr-1 can be used to calculate the total number of ways it can be done. Blanks are permitted in this distribution, which implies a person could also receive nothing.

If we were to split n identical objects among r people in such a way that each individual receives 1, 2, or more items until the objects were >0 and ≤n, the number of possible ways to do so might be calculated using the formula n-1Cr-1. Blanks are not permitted in this distribution, therefore each participant must receive at least one item.

If we need to divide n identical things into r groups so that no group receives fewer than k items or more than m items such that (mk), then the coefficient of xn in the expansion of (xm+xm+1+….+xk)r will tell us how many different ways we can do it.

If there are n objects put in a row, then the number of arrangements that may be made so that none of the objects takes up their original positions is given by D(n), which equals

n! {1−1/1!+1/2!−1/3!+1/4!−….+(−1)n1/n!}.

If we assume that there are r objects and that 0≤r≤n holds, all of the objects occupy their original locations, and none of the remaining items does, then this can be done in a variety of ways, including the following;

D(n−r)=nCr.D(n−r) =nCr.(n−r)

! {1−1/1!+1/2!−1/3!+1/4!−….+(−1)n−r1/(n−r)!}