NCERT Solutions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry

March 23, 202339 Insightful Publications

**Permutation and Combination:** Permutation and Combination are one of the most important concepts in Mathematics. It is used in the practical applications in science, engineering and research, and many other fields. The difference between permutation and combination can be defined as; when a set of data is selected from a certain group, it is known as permutation; while the order in which the data is arranged is known as combination. Thus, we can say that permutation is an ordered combination.

Grouping of data and its interpretation is possible with the help of permutation and combination. Mathematics students should be well versed in the concepts because it finds applications in all advanced research and data analysis. This article provides insight on the permutation and combination questions, solutions, permutation and combination formulas and more.

Here we have given the mathematical definition of permutation and combination:

**What is Permutation?**

A permutation is an arrangement in a definite order of several objects taken, some or all at a time, with permutations, every tiny detail matters. It means the order in which elements are arranged is significant.

There are two types of permutations:

**Repetition is Allowed:**For the number lock example provided above, it could be “2-2-2”.**No Repetition Allowed:**For example, the first three people in a race. You can’t be first and second at the same time.

**What is Combination?**

The combination is a way of selecting elements from a set so that the order of selection doesn’t matter. With the combination, only choosing elements matters. It means the order in which elements are chosen is not essential.

There are two types of combinations:

**Repetition is Allowed**: For example, coins in your pocket (2, 5, 5, 10, 10)**No Repetition Allowed**: For example, lottery numbers (2, 14, 18, 25, 30, 38)

There are many formulas that are used to solve permutation and combination problems. We have provided the complete permutation formula list here:

**When repetition is not allowed:**P is a permutation or arrangement of r things from a set of n things without replacement.**When repetition is allowed:**P is a permutation or arrangement of r things from a set of n things when duplication is allowed.

**Derivation of Permutation Formula:** Let us assume that there are r boxes, and each of them can hold one thing. There will be as many permutations as there are ways of filling in *r *vacant boxes by *n* objects.

- No. of ways the first box can be filled:
*n*

- No. of ways the second box can be filled:
**(***n*– 1)

- No. of ways the third box can be filled:
**(****n****– 2)**

- No. of ways the fourth box can be filled:
**(****n****– 3)**

- No. of ways r
^{th}box can be filled:**[***n*– (*r*– 1)]

- The number of permutations of
*n*different objects taken*r*at a time, where 0 <*r ≤ n*and the objects do not repeat is:**n****(****n****– 1)(****n****– 2)(****n****– 3) . . . (****n****– r + 1)**

We have provided the complete combination formula list here:

**When repetition is not allowed:**C is a combination of n distinct things taking r at a time (order is not important).**When repetition is allowed:**C is a combination of n distinct things taking r at a time (order is not important) with repetition. We define C as:

**Derivation of Combination Formula:**

Let us assume that there are r boxes, and each of them can hold one thing.

- No. of ways to select the first object from
*n*distinct objects:**n** - No. of ways to select the second object from (
*n-1)*distinct objects:**(****n-1)** - No. of ways to select the third object from (
*n-2)*distinct objects:**(****n-2)** - No. of ways to select r
^{th}object from [*n-(r-1)*] distinct objects:**[****n-(r-1)****]**

Completing the selection of r things from the original set of n things creates an ordered subset of * r* elements.

∴ The number of ways to make a selection of r elements of the original set of

Let us consider the ordered subset of *r* elements and all their permutations. The total number of permutations of this subset equals r! because *r *objects in every combination can be rearranged in *r! *ways.

Hence, the total number of permutations of *n *different things taken *r *at a time is (nC_{r}×r!). It is nothing but nP_{r}.

**Download** – **Permutation and Combination Formula PDF**

We can summarize the permutation combination formula in the table below:

We have provided the permutation and combination differences in the table below:

Permutation | Combination |

A selection of r objects from a set of n objects in which the order of the selection matters. | The number of possible combinations of r objects from a set on n objects where the order of selection doesn’t matter. |

A permutation is used for lists (order matters). | The combination is used for groups (order doesn’t matter). |

It denotes the arrangement of objects. | It does not denote the arrangement of objects. |

We can derive multiple permutations from a single combination. | Only a single combination can be derived from a single permutation. |

They are defined as ordered elements. | They are defined as unordered sets. |

**Check:**

Algebra Formulas for Class 8 | Algebra Formulas for Class 9 |

Algebra Formulas for Class 10 | Algebra Formulas for Class 11 |

We have provided some permutation and combination examples with detailed solutions. Get **Permutation and Combination Class 11 NCERT Solutions** for free on Embibe.

**Q.1: Find the number of permutations and combinations, if n = 15 and r = 3.****Ans:** n = 15, r = 3 (Given)

Using the formulas for permutation and combination, we get:

Permutation, P = n!/(n – r)!

= 15!/(15 – 3)!

= 15!/12!

= (15 x 14 x 13 x 12!)/12!

= 15 x 14 x 13

= 2730

Also, Combination, C = n!/(n – r)!r!

= 15!/(15 – 3)!3!

= 15!/12!3!

= (15 x 14 x 13 x 12! )/12!3!

= 15 x 14 x 13/6

= 2730/6

= 455

**Q.2: In how many different ways can the letters of the word ‘OPTICAL’ be arranged so that the vowels always come together? ****Ans:** The word ‘OPTICAL’ has 7 letters. It has the vowels’ O’, ‘I’, and ‘A’ in it and these 3 vowels should always come together. Hence these three vowels can be grouped and considered as a single letter. That is, PTCL(OIA).

Hence we can assume the total letters as 5 and all these letters are different.

Number of ways to arrange these letters

= 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

Hence, the required number of ways:

= 120×6

= 720

**Q.3: How many 3-letter words with or without meaning, can be formed out of the letters of the word, ‘LOGARITHMS’ if repetition of letters is not allowed? ****Ans:** The word ‘LOGARITHMS’ has 10 different letters.

Hence, the number of 3-letter words (with or without meaning) formed by using these letters

= ^{10}P_{3}

= 10×9×8

= 720

**Q.4: There are 8 men and 10 women and you need to form a committee of 5 men and 6 women. In how many ways can the committee be formed? ****Ans:** We need to select 5 men from 8 men and 6 women from 10 women.

Number of ways to do this

= ^{8}C_{5} × ^{10}C_{6}

= ^{8}C_{3} × ^{10}C_{4} [∵ ^{n}C_{r} = ^{n}C_{(n-r)}]

= [(8 x 7 x 6)/(3 x 2 x 1)] x [(10 x 9 x 8 x 7)/(4 x 3 x 2 x 1)]

= 56×210

= 11760

**Q.5: A bag contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the bag if at least one black ball is to be included in the draw? ****Ans:** From 2 white balls, 3 black balls and 4 red balls, 3 balls are to be selected such that at least one black ball should be there.

Hence we have 3 choices as given below.**Choice 1:** We can select 3 black balls.**Choice 2:** We can select 2 black balls and 1 non-black ball.**Choice 3:** We can select 1 black ball and 2 non-black balls.

Number of ways to select 3 black balls

= 3C3

Number of ways to select 2 black balls and 1 non-black ball

= 3C2 × 6C1

Number of ways to select 1 black ball and 2 non-black balls

= 3C1 × 6C2

Total number of ways

= 3C3 + 3C2 × 6C1 + 3C1 × 6C2

= 3C3 + 3C1 × 6C1 + 3C1 × 6C2[∵ ^{n}C_{r} = ^{n}C_{(n-r)}]

= 1 + (3×6) + [3 x (6×5)/(2×1)]

= 1 + 18 + 45

= 64

**Q.6: An event manager has ten patterns of chairs and eight patterns of tables. In how many ways can he make a pair of tables and chairs? ****Ans:** The event manager has 10 patterns of chairs and 8 pattern 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 × 8 ways

= 80 ways

**Q.7: In how many ways can three boys can be seated on five chairs? ****Ans:** There are 3 boys.

The 1st boy can sit in any of the five chairs (5 ways).

5 |

Now there are 4 chairs remaining. The second boy can sit in any of the four chairs (4 ways).

5 | 4 |

Now there are 3 chairs remaining. The third boy can sit in any of the three chairs (3 ways).

5 | 4 | 3 |

Hence, the total number of ways in which 3 boys can be seated on 5 chairs

= 5 × 4 × 3

= 60

**Q.8: In how many ways can a team of 5 persons be formed out of a total of 10 persons such that two particular persons should be included in each team? **

Answer: Two particular persons should be included in each team. Therefore we have to select the remaining (5 – 2) = 3 persons from (10 – 2) = 8 persons.

Hence, the required number of ways

= ^{8}C_{3}

= (8×7×6)/(3×2×1)

= 8×7

= 56

*Other important Maths articles:*

Here are some practice questions on permutation and combination concepts for you to practice:

Question 1: 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 number of different arrangements possible. Question 2: In how many ways can a team of 5 persons be formed out of a total of 10 persons such that two particular persons should not be included in any team?Question 3: If there are 9 horizontal lines and 9 vertical lines on a chessboard, how many rectangles can be formed on the chessboard? Question 4: Find the number of triangles that can be formed using 14 points in a plane such that 4 points are collinear?Question 5: What is the sum of all 4 digit numbers formed using the digits 2, 3,4 and 5 without repetition?Question 6: At a birthday party, every person shakes hands with every other person. If there was a total of 28 handshakes at the party, how many persons were present at the party?Question 7: If ^{n}C_{8} = ^{n}C_{27}, what is the value of n?Question 8: Find the number of triangles which can be drawn out of n given points on a circle?Question 9: In how many ways can 10 books be arranged on a shelf such that a particular pair of books will never be together? Question 10: How many numbers, between 100 and 1000, can be formed with the digits 3, 4, 5, 0, 6, 7 (repetition of digits is not allowed)? |

*You can also check,*

Q.1: What is nPr formula? nPr = n!/(n – r)!Ans: |

Q.2: What is the formula for permutation and combination?Ans: The number of permutations of n objects taken r at a time is determined by the following formula: P(n,r) = n!/(n−r)!. The number of combinations of r objects from a set of n objects is determined by the following formula: C(n,r) = n!/(n−r)!r! |

Q.3: What is the use of permutation and combination in real life?Ans: There are many uses of permutation and combination in our day-to-day life. Some common areas where permutation is used are arranging people, digits, numbers, alphabets, letters, and colours. Selection of menu, food, clothes, subjects, the team are examples of combinations. |

Q.4: What is nCr formula?Ans: nCr = n! /(n – r)!r! |

Q.5: Where do we use permutation and combination?Ans: We use permutation for a list of objects where the order of selection matters. On the other hand, the combination is used for a group of objects where the order of selection is irrelevant. |