## Revision Notes for CBSE Class 11 Maths Chapter 6 (Permutations and Combinations) - Free PDF Download

There are some chapters that students always have difficulty with. Permutations and Combinations is one of those chapters in Mathematics. This is why it is suggested that students should go through Permutation And Combination Class 11 Notes by Vedantu. If you want to do just that, then you are in the right place. In these Class 11 Permutation And Combination Notes, we will talk about all the significant points that were included in this chapter. So, let’s start with the basics. According to the Class 11 Maths Ch 6 Notes PDF, permutation and combination can be explained as methods for representing a group of objects. You can also opt to download the PDF of Permutation and Combination Notes for practice or revision purposes.

## Download CBSE Class 11 Maths Revision Notes 2024-25 PDF

Also, check CBSE Class 11 Maths revision notes for all chapters:

CBSE Class 11 Maths Chapter-wise Notes | |

Chapter 7 Permutations and Combinations Notes | |

Permutations and Combinations Chapter-Related Important Study Materials

It is a curated compilation of relevant online resources that complement and expand upon the content covered in a specific chapter. Explore these links to access additional readings, explanatory videos, practice exercises, and other valuable materials that enhance your understanding of the chapter's subject matter.

Permutations and Combinations Related Other Study Materials |

## Permutations and Combinations Class 11 Notes Maths - Basic Subjective Questions

### Section–A (1 Mark Questions)

1. If ${ }^n C_{10}={ }^n C_{12}$ then find n.

Ans. We have, ${ }^n C_{10}={ }^n C_{12}$

We know that,

$$ \begin{aligned} & { }^n C_x={ }^n C_y \Rightarrow x=y \text { or } n=x+y . \\ & \text { Since, } 10 \neq 12 \Rightarrow 10+12=n \\ & \Rightarrow n=22 . \end{aligned} $$

2. How many four lettered words can be made using the letters of the word ‘ROCKY’.

Ans. The word ROCKY consists of 5 letters.

In order to make four letter words, we need to permute these 5 letter taken four at time.

Therefore,

Required number of words

$$={ }^5 P_4=\frac{5 !}{(5-4) !}=\frac{5 !}{1 !}=120 \text {. }$$

3. Number of ways of selecting 2 students for a drive out of 8 students is.

Ans. Number of ways of selecting 2 students for a drive out of 8 students

$={ }^8 C_2=\frac{8 !}{2 !(8-2) !}=\frac{8 \times 7 \times 6 !}{2 \times 6 !}=28 \text {. }$

4. Find the number of triangles formed using 8 points on the circle.

Ans. Number of triangles formed is equal to number of selections that can be made by selecting 3 objects from 8 distinct objects which is equal to

$${ }^8 C_3=\frac{8 !}{3 !(8-3) !}=\frac{8 \times 7 \times 6 \times 5 !}{3 ! \times 5 !}=56\text {. }$$

5. How many different parties of 4, consisting of 1 boy and 3 girls, can be made from 5 boys and 5 girls.

Ans. Clearly, out of the 5 boys and 5 girls, 1 boy and 3 girls will be chosen, Then, number of different parties

$$ \begin{aligned} & ={ }^5 C_1 \times{ }^5 C_3 \\ & =5 \times \frac{5 \times 4 \times 3}{6} \\ & =50 . \end{aligned} $$

### Section–B (2 Marks Questions)

6. If ${ }^{15} C_{3 r}={ }^{15} C_{r+3}$ find r.

Ans. Given: ${ }^{15} C_{3 r}={ }^{15} C_{r+3}$

We know that,

$$ \begin{aligned} & { }^n C_x={ }^n C_y \Rightarrow x=y \text { or } n=x+y \\ & \Rightarrow 3 r=r+3 \text { or, } 3 r+r+3=15 \\ & \Rightarrow 2 r=3 \text { or, } 4 r=12 \\ & \Rightarrow r=\frac{2}{3} \text { or, } r=3 \\ & \Rightarrow r=3 \quad\left\{\because r \neq \frac{2}{3}\right\} . \end{aligned} $$

7. Find r, if $5{ }^4 P_r=6{ }^5 P_{r-1}$ .

Ans. We have $5{ }^4 P_r=6{ }^5 P_{r-1}$

$$ \begin{array}{ll} \text { Or } & 5 \times \frac{4 !}{(4-r) !}=6 \times \frac{5 !}{(5-r+1) !} \\ \text { Or } & \frac{5 !}{(4-r) !}=\frac{6 \times 5 !}{(5-r+1)(5-r)(5-r-1) !} \\ \text { Or } \quad(6-r)(5-r)=6 \\ \text { Or } \quad r^2-11 r+24=0 \\ \text { Or } \quad r^2-8 r-3 r+24=0 \\ \text { Or } \quad(r-8)(r-3)=0 \\ \text { Or } \quad r=8 \quad \text { or } r=3 . \\ \text { Hence, } \quad r=8,3 . \end{array} $$

8. If $\frac{1}{8 !}+\frac{1}{9 !}=\frac{x}{10 !}$ find x.

Ans. We have,

$$ \begin{aligned} & \frac{1}{8 !}+\frac{1}{9 !}=\frac{x}{10 !} \\ & \Rightarrow \frac{1}{8 !}+\frac{1}{9 \times 8 !}=\frac{x}{10 \times 9 \times 8 !} \\ & \Rightarrow 1+\frac{1}{9}=\frac{x}{10 \times 9} \\ & \Rightarrow \frac{10}{9}=\frac{x}{10 \times 9} \\ & \Rightarrow x=100 . \end{aligned} $$

9. Find the number of diagonals in an octagon.

Ans. Octagon has 8 corner points,

Therefore, Number of lines joining any two of these eight points ${ }^{8} C_{2}=\frac{8\times7}{2}=28$

Out of these 28 lines, 8 are the sides of octagon and rest are diagonals.

Thus, Number of diagonals =28-8=20.

10. Convert $3\times6\times9\times12\times15\times18$ into factorials.

Ans. $3\times6\times9\times12\times15\times18$

$=(3\times1)\times(3\times2)\times(3\times3)\times(3\times4)\times(3\times5)\times(3\times6)$

=36(1\times2\times3\times4\times5\times6)

=36(6!)

11. There are 10 points in a plane of which 4 are collinear. How many different straight lines can be drawn by joining these points.

Ans. Number of straight lines formed joining the 10 points, taking 2 points at a time $={ }^{10} C_{2}=\frac{10}{2}\times\frac{9}{1}=45$

Number of straight lines formed joining the 4 points, taking 2 points at a time $={ }^{4} C_{2}=\frac{4}{2}\times\frac{3}{1}=6$

But, when 4 collinear points are joined pair wise, they give only one line.

Required number of straight lines = 45-6+1+40.

12. Find the number of 4 cards combinations in which 2 are black and 2 are red out of a deck of 52 card.

Ans. Since, there are 26 black and 26 red cards in a deck of 52 cards.

Therefore,

Number of ways of selecting 2 black cards $={ }^{26} C_{2}=\frac{26\times 25}{2}=325$ ways

Number of ways of selecting 2 red cards $={ }^{26} C_{2}=\frac{26\times 25}{2}=325$ ways

Thus by multiplication principle,

Required number of combinations $= 325\times325=105625$.

13. Find the 3-digits number greater than 300 formed by using digits 0, 2, 3, 5, 7 without repetition.

Ans. Since, the number is greater than 300 and repetition is not allowed.

Hundred’s place can be occupied by 3 ways {3,5,7}.

Once the hundred’s place is occupied, we are left with 4 numbers to be arranged at 2 places which can be done in ${ }^{4} P_{2}=\frac{4!}{2!}=12$ ways.

Thus, by multiplication principle.

Total numbers greater than 300 = $3\times12$=36.

PDF Summary - Class 11 Maths Permutations and Combinations Notes (Chapter 6)

Fundamental Principles Of Counting:

**Fundamental principle of multiplication –**If there are three different events such that one event occurs in $m$ different ways, second event happens in $n$ different ways and the third event occurs in $p$ different ways, then all three events simultaneously will happen in $m\times n\times p$ ways.**Fundamental principle of addition –**If there are two jobs such that first wone can be performed independently in $m$ number of ways and the second work independently can be done in $n$ number of ways, then wither of the two jobs can be performed in $\left( m+n \right)$ ways.

Some Basic Arrangements and Selections:

Permutations – It is the linear arrangements of distinct objects taken some or all at a time. The number of arrangements possible is called the permutations. If we have two positive integers $r$ and $n$ such that $l\le r\le n$, then the total number of arrangements or permutations possible for $n$ distinct items taken $r$ at a time is mathematically given by,

${}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$.

Combinations – If we have to select combinations of items from a given set of items such that the order or arrangement doesn’t matter, then we use combinations. Such that to find the number of ways of selecting $r$ objects from a set of $n$ objects, then mathematically it is given by,

${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$

Geometric Applications Of ${}^{n}{{C}_{r}}$:

If there are $n$ non-parallel and non-concurrent lines, then the number of points of intersections is given by ${}^{n}{{C}_{2}}$. Similarly, having $n$ points, the number of lines will be ${}^{n}{{C}_{2}}$ such that no three points are collinear.

If in $n$ number of points, there are $m$ collinear, then number of straight lines is given by ${}^{n}{{C}_{2}}-{}^{m}{{C}_{2}}+1$.

If a polygon has $n$ number of vertices such that no three are collinear, then the number of diagonals can be given by ${}^{n}{{C}_{2}}-n=\dfrac{n\left( n-3 \right)}{2}$.

If we have $n$ points such that no three are collinear, then the number of triangles that can be formed is given by ${}^{n}{{C}_{3}}$.

If we have $n$ points out of which $m$ are collinear, then the number of triangles that can be formed is given by ${}^{n}{{C}_{3}}-{}^{m}{{C}_{3}}$.

A polygon having $n$ vertices, then the number of triangles that can be formed such that no side of the triangle is common to that of the side of the polygon is given by ${}^{n}{{C}_{3}}-{}^{n}{{C}_{1}}-\left( {}^{n}{{C}_{1}}\times {}^{n-4}{{C}_{1}} \right)$.

Number of parallelograms that can be formed from two sets of parallel lines such that one set have $n$ parallel lines and other set have $m$ parallel lines is given by ${}^{n}{{C}_{2}}\times {}^{m}{{C}_{2}}$.

Number of squares that can be formed from two sets of equally spaced parallel lines such that one set have $n$ parallel lines and other set have $m$ parallel lines is given by $\underset{r-1}{\overset{m-1}{\mathop \sum }}\,\left( m-r \right)\left( n-r \right);\left( m<n \right)$.

Permutations Under Certain Conditions:

The total number of arrangements or permutations taken $r$ at a time from a set of $n$ different objects;

When we always have to include a particular object in every arrangement is ${}^{n-1}{{C}_{r-1}}\times r!$.

When we don’t have to include a particular object in any arrangement it is ${}^{n-1}{{C}_{r}}\times r!$.

Division Of Objects Into Groups:

Division Of Items Into Groups Of Unequal Sizes -

A set of $\left( m+n \right)$ distinct objects can be divided into two unequal groups containing $m$ and $n$ objects is given by $\dfrac{\left( m+n \right)!}{m!n!}$.

A set of $\left( m+n+p \right)$ distinct objects can be divided into three unequal groups containing $m,n$ and $p$ objects is given by ${}^{m+n+p}{{C}_{m}}.{}^{n+p}{{C}_{m}}=\dfrac{\left( m+n+p \right)!}{m!n!p!}$.

A set of $\left( m+n+p \right)$ distinct objects can be divided into three persons in the groups containing $m,n$ and $p$ objects is

$=(\text{No}\text{. of ways to divide})\times (\text{No}\text{. of groups})!$

$=\dfrac{\left( m+n+p \right)!}{m!n!p!}\times 3!$.

Divisions Of Objects Into Groups Of Equal Sizes –

We can divide $mn$ distinct objects equally into $m$ groups with $n$ objects in each of them without considering any order in $\left( \dfrac{\left( mn \right)!}{{{\left( n! \right)}^{m}}} \right)\dfrac{1}{m!}$ ways.

We can divide $mn$ distinct objects equally into $m$ groups with $n$ objects in each of them considering the order in $\left( \left( \dfrac{\left( mn \right)!}{{{\left( n! \right)}^{m}}} \right)\dfrac{1}{m!} \right)m!=\dfrac{\left( mn \right)!}{{{\left( n! \right)}^{m}}}$ ways.

Permutations Of Alike Objects:

Total number of mutually distinguishable arrangements or permutations possible for $p+q=n$ objects among which $p$ objects are of one kind and $q$ are alike of second kind taken all at a time is given by $\dfrac{n!}{p!q!}$.

Total number of mutually distinguishable arrangements or permutations possible for $p+q\ne n$ objects among which $p$ objects are of one kind and $q$ are alike of second kind and remaining all are distinct taken all at a time is given by $\dfrac{n!}{p!q!}$.

Total number of mutually distinguishable arrangements or permutations possible for ${{p}_{1}}+{{p}_{2}}+{{p}_{3}}+.....+{{p}_{r}}=n$ objects such that, ${{p}_{1}}$ are alike of one kind; ${{p}_{2}}$ are alike of second kind;……; ${{p}_{r}}$ are alike of ${{r}^{th}}$ kind taken all at a time is given by $\dfrac{n!}{{{p}_{1}}!{{p}_{2}}!{{p}_{3}}!......{{p}_{r}}!}$.

If we have to arrange $r$ things such that the repetitions are allowed. Furthermore, if ${{p}_{1}},{{p}_{2}},{{p}_{3}},.....,{{p}_{r}}$ are integers such that it denotes the number of times respective objects occurs, then the total number of permutations of these $r$ objects considering above condition is given by $\dfrac{\left( {{p}_{1}}+{{p}_{2}}+{{p}_{3}}+.....+{{p}_{r}} \right)!}{{{p}_{1}}!{{p}_{2}}!{{p}_{3}}!......{{p}_{r}}!}$.

Distribution Of Alike Objects:

Suppose, we have to divide x$n$ identical items among $r$ persons, such that each person receives $0,1,2$ or more items until it is $\le n$, then the number of ways it can be done can be given by ${}^{n+r-1}{{C}_{r-1}}$. In this distribution, blanks are allowed means a person may also get no items.

Suppose, we have to divide $n$ identical items among $r$ persons, such that each person receives $1,2,3$ or more items until it is $>0$ and $\le n$, then the number of ways it can be done can be given by ${}^{n-1}{{C}_{r-1}}$. In this distribution, blanks are not allowed means a person has to get at least one item.

Suppose, we have to divide $n$ identical items among $r$ groups, such that no group get less than $k$ and more than $m$ items such that $\left( m<k \right)$, then the number of ways it can be done is the coefficient of ${{x}^{n}}$ in the expansion of ${{\left( {{x}^{m}}+{{x}^{m+1}}+....+{{x}^{k}} \right)}^{r}}$.

Number Of Integral Solutions Of Linear Equations And Inequation:

Let us consider the equation ${{x}_{1}}+{{x}_{2}}+{{x}_{3}}+{{x}_{4}}+.....+{{x}_{r}}=n$, where ${{x}_{1}},{{x}_{2}},{{x}_{3}},{{x}_{4}},.....,{{x}_{r}}$ and $n$ are non-negative integers. We can interpret this equation as $n$ identical objects to be divided into $r$ groups.

The total number of non-negative integral solutions of the equation ${{x}_{1}}+{{x}_{2}}+{{x}_{3}}+{{x}_{4}}+.....+{{x}_{r}}=n$ will be ${}^{n+r-1}{{C}_{r-1}}$.

In the set of natural numbers $N$, the total number of solutions of the equation will be ${}^{n-1}{{C}_{r-1}}$.

We can solve the equations of the form ${{x}_{1}}+{{x}_{2}}+{{x}_{3}}+{{x}_{4}}+.....+{{x}_{m}}\le n$ by introducing an artificial or dummy variable ${{x}_{m+1}}$ such that ${{x}_{m+1}}\ge 0$. It will convert the equation to ${{x}_{1}}+{{x}_{2}}+{{x}_{3}}+{{x}_{4}}+.....+{{x}_{m}}+{{x}_{m+1}}=n$ and then the number of solutions can be found out in the same way as given in above points.

Circular Permutations:

If we have $n$ distinct objects, then the number of circular permutations possible is given by $\left( n-1 \right)!$.

If in circular permutations, we also consider the anti-clockwise and clockwise order as non-distinct, then the number of circular permutations can be given by $\dfrac{1}{2}\left\{ \left( n-1 \right)! \right\}$. This can be seen in the arrangement of beads in a necklace or flowers in a garland, etc.

Selection Of One Or More Objects:

If we have a group of $n$ distinct items, then the number of ways in which we can select one or more items from that group is given by ${{2}^{n}}-1$.

If we have a group of $n$ identical items, then the number of ways in which we can select $r$ items from that group is always $1$.

If we have a group of $n$ identical items, then the number of ways in which we can select zero or more items from that group is $n+1$.

If we have a group of $\left( p+q+r \right)$ items, such that $p$ items are alike of one kind, $q$ items are alike of second kind and $r$ items are alike of third kind, then the number of ways in which we can select some or all out of given items is given by $\left[ \left( p+1 \right)\left( q+1 \right)\left( r+1 \right) \right]-1$.

If we have a group of $\left( p+q+r \right)$ items, such that $p$ items are alike of one kind, $q$ items are alike of second kind and $r$ items are alike of third kind, then the number of ways in which we can select some or all out of given items is given by $\left[ \left( p+1 \right)\left( q+1 \right)\left( r+1 \right) \right]-1$.

If we have a group of $\left( p+q+r \right)$ items, such that $p$ items are alike of one kind, $q$ items are alike of second kind and $r$ items are alike of third kind, then the number of ways in which we can select one or more items is given by $\left[ \left( p+1 \right)\left( q+1 \right)\left( r+1 \right){{2}^{n}} \right]-1$.

Number of Divisors And The Sum Of The Divisors Of A Given Natural Number:

Let us consider $N={{p}_{1}}^{{{n}_{1}}}.{{p}_{2}}^{{{n}_{2}}}.{{p}_{3}}^{{{n}_{3}}}.{{p}_{4}}^{{{n}_{4}}}......{{p}_{k}}^{{{n}_{k}}}$, where ${{p}_{1}},{{p}_{2}}....,{{p}_{k}}$ are distinct prime numbers and ${{n}_{1}},{{n}_{2}}....,{{n}_{k}}$ are positive integers.

Then, the total number of divisors $=\left( {{n}_{1}}+1 \right)\left( {{n}_{2}}+1 \right)...\left( {{n}_{k}}+1 \right)$.

The above divisors also include $1$ and $n$ as divisors, then the number of divisors other than $1$ and $n$ is $=\left( {{n}_{1}}+1 \right)\left( {{n}_{2}}+1 \right)...\left( {{n}_{k}}+1 \right)-2$.

And the sum of all divisors is given by

$=\left\{ \dfrac{{{p}_{1}}^{{{n}_{1}}+1}}{{{p}_{1}}-1} \right\}\left\{ \dfrac{{{p}_{2}}^{{{n}_{2}}+1}}{{{p}_{2}}-1} \right\}\left\{ \dfrac{{{p}_{3}}^{{{n}_{3}}+1}}{{{p}_{3}}-1} \right\}....\left\{ \dfrac{{{p}_{k}}^{{{n}_{k}}+1}-1}{{{p}_{k}}-1} \right\}$.

Dearrangements:

Suppose $n$ objects are arranged in a row, then the number of ways in which we can arrange the objects so that none of them occupies their original places is given by $D(n)$ which is equal to

$n!\left\{ 1-\dfrac{1}{1!}+\dfrac{1}{2!}-\dfrac{1}{3!}+\dfrac{1}{4!}-....+{{(-1)}^{n}}\dfrac{1}{n!} \right\}$.

Suppose we have $r$ number of objects such that $0\le r\le n$ and each object occupies their original places assigned to them and none of the remaining objects occupies their original places, then this can be done in following number of ways;

$D\left( n-r \right)={}^{n}{{C}_{r}}.D\left( n-r \right)$

\[={}^{n}{{C}_{r}}.\left( n-r \right)!\left\{ 1-\dfrac{1}{1!}+\dfrac{1}{2!}-\dfrac{1}{3!}+\dfrac{1}{4!}-....+{{(-1)}^{n-r}}\dfrac{1}{\left( n-r \right)!} \right\}\].

## Revision Notes CBSE Class 11 Maths Notes Chapter 6 Permutations and Combinations

## What are Permutations and Combinations?

Permutations and combinations define the different ways in which a particular group of data can be arranged. This is done by selecting the items in a set or by forming subsets. If we look at the specific definitions, then permutation is when one selects the objects or data from a specific group. On the other hand, the combination is the order in which the data is represented. Both of these concepts play a vital role in the field of mathematics. In these Permutations And Combinations Class 11 Notes, we will get into the topic and explore the various aspects that are related to this topic.

Permutation can also be described as the phenomenon of rearranging the elements of data that were already present in order before. If we look at the broad field of Mathematics, then it is easy to say that permuting happens in almost all areas of this subject. Permuting is often more common when an individual wants to consider ordering data in a different manner. This is done on a specific finite set.

Combination is just the manner of selecting items from a collection. In this practice, the order of selection is not considered to be of any significant consequence. It is also possible for individuals to calculate the total number of combinations. It would help if you remembered that this is true for smaller cases only. Combination can also be defined as the combination of n things that are taken k at a time. This should be done without any repetition. In cases of combinations where repetition is allowed, the terms k-selection or k-combination with repetition are usually used.

It is advised that all students should go through Permutation And Combination Notes Class 11 as these notes can help students to score the highest marks in their respective classes. There are also many formulas that are related to the concepts of combination and permutation. However, there are two significant formulas that are used in most cases. These formulas are:

### Permutation Formula

One can define permutation as the choice of r things taken from a set of n things. This is done without any replacement. Also, in the case of permutation, the order of things will matter. All of this can be represented by the formula mentioned below.

nPr = (n!) / (n - r)!

### Combination Formula

A combination can be explained as the choice of r things that are made from a set of n things. In the case of combination, all of this is done without replacement, and the order of things does not matter. All of this information can be represented by the formula:

nCr = nPr / r! = n! / r! (n - r) !

(r)

(Image to be added soon)

### Difference Between Permutation and Combination

By now, readers must be able to understand the basic meaning of permutation and combination with the help of these notes of permutation and combination. This is why it is now time to look at the difference that exists between both of these concepts.

Further, to help students better understand this topic, we have formulated a table that contains all the points related to the difference between combination and permutation. That table is mentioned below.

Permutation | Combination |

It refers to the task of arranging digits, people, alphabets, colours, numbers, and letters | It is the selection of food, menu, clothes, teams, subjects, and other objects |

An example of permutation is to pick a team captain, picture, or shortstop from a group | An example of combinations includes picking any three team members from a group |

Deciding on your two favourite colours in a particular order from a colour brochure | Selecting any two colours from a colour brochure |

Picking winners for the first, second, and third place | Picking any three winners for an award |

### Fun Facts about Permutation and Combination

Did you know permutations and combinations have various uses in our daily lives as well as academic lives? We can all use permutation and combination to make our lives easier.

For example, if you have to work on solving questions in the order of their difficulty or when you have to pick colours to paint a room. These applications will help students to understand these class 11 maths chapter 6 notes in a more effective manner.

## Why should you Choose Veadntu for Class 11 Chapter 6 - Permutations and Combinations for Revision Notes?

1. Clear Summaries: Quickly grasp key concepts for efficient learning.

2. Simplified Topics: Enhance understanding with simplified complex topics.

3. Last-Minute Prep: An efficient tool for effective last-minute exam preparation.

4. Retention Aid: Improve information retention for better recall during exams.

5. Exam Support: Get ready for exams with essential points and tips.

6. Time Saver: Save time by accessing consolidated information in one place.

7. Priority Focus: Focus on important topics and questions for targeted preparation.

8. Real-World Examples: Practical examples for better real-world connections.

9. Confidence Booster: Boost confidence for exams with comprehensive notes.

## Conclusion

The Permutations and Combinations Class 11 Notes for CBSE Maths Chapter 6 offer a comprehensive understanding of key concepts. Students gain valuable insights into the properties and characteristics of permutations and combinations. The well-organized format facilitates efficient revision, ensuring better retention. Practical examples and real-life applications enrich the subject's understanding. These notes are essential for establishing a robust foundation in permutations and combinations, paving the way for academic excellence. Proven invaluable, these notes significantly contribute to students' success in their studies. Access the free PDF download for a thorough grasp of this vital chapter.

## FAQs on Permutations and Combinations Class 11 Notes CBSE Maths Chapter 6 [Free PDF Download]

1. Find the total number of permutations and combinations if the value of n is 12 and the value of r is 2.

We know that n = 12 and r = 2

By using the formula for permutation, we will get:

nPr = (n!) / (n - r)! = (12!) / (12 - 2)! = 12! / 10! = (12 x 11 x 10!) / 10! = 132

If we use the formula for combination, then we will get:

nCr = n1 / r! (n - 1)! . 12! / 2! (12 - 2)! = 12! / 2! (10)! = 12 x 11 x 10! / 2! (10)! = 66

2. You know that in a dictionary all permutations of the letters that are used in the word AGAIN are arranged in a particular order. Find out the 49th word by using this information.

Let’s start with the letter A. If we arrange the other four letters, then we get G, A, I, N - 4! = 24. These are the first 24 words.

Now, let’s start with the letter G. If we arrange it as A, A, I, and N in various ways, then we will get 4! / 2! = 12. These are the next 12 words.

Finally, let’s begin with the letter I. If we arrange the letters in the order of A, A, G, and N in various ways, then we will get 4! / 2! = 12. There are the next 12 words.

Hence, 24 + 12 + 12 = 48. These are the 48th word. This is why we can say that the 49th word is ‘NAAGI.’

3. There is a group of nine men and twelve women. Calculate the number of ways in which a committee of five men and three women can be chosen from the given group.

If we choose five men out of nine men = ^{9}C_{5} ways = 126 ways

If we choose 3 women out of 12 women = ^{12}C_{3} ways = 220 ways

Hence, the committee can be chosen in 27720 ways!

4. Define permutation and combination.

Permutation can be defined as the process of arranging numbers or objects in a particular order. Combination, on the other hand, can be defined as the method of picking items or numbers from a collection of things in a manner in which the order of the objects is not a significant consideration.

5. What are the formulas for calculating permutations and combinations?

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

The formula for calculating combinations is: nCr = n! / [r! (n - r)!]

6. In what manner are both permutations and combinations related to one another?

The formula that shows the relationship between permutations and combinations is:

nCr = nPr / r!