## Permutation and Combination

We have always faced the problem of understanding the difference between permutation and combination.This chapter will definitely clear the concepts of permutation and combination, the only thing you have to do is thoroughly understand the difference between the two terms and as well learn the quick tips to solve problems based on this chapter

**Difference between permutation and combination**

**What is permutation?**

**Permutation:**The various ways of arranging a given number of things by taking some or all at a time are all called as permutations.

Permutation includes word formation, number formation, circular permutation, etc. In permutation, objects are to be arranged in particular order. It is denoted by

^{n }P

_{r }or P(n, r).

**Example:**Arrange the given 3 numbers 1, 2, 3 by taking two at a time.

Now these numbers can be arranged in 6 different ways: (12, 21, 13, 31, 23, 32).

Here,

12 and 21, 13 and 31 or 23 and 32 do not mean the same, because here order of numbers is important.

**What is combination?**

**Combination:**Each of different groups or selections formed by taking some or all number of objects is called a combination.

Combination is used in different cases which include team/group/committee.

In combination, objects are selected randomly and here order of objects doesn’t matter. It is denoted by

^{n }C_{r }or C(n, r)**Example:**If we have to select two girls out of 3 girls X, Y, Z, then find the number of combinations possible.

Now only two girls are to be selected and arranged. Hence, this is possible in 3 different ways: (XY, YZ, XZ,).

Here,

You cannot make a combination as XY and YX, because these combinations mean the same.

**Quick Tips and Tricks**

**Factorial n!:**It is the product of all positive integers less than or equal to n.

Example: 4! = 4 × 3 × 2 × 1 = 24

**Theorem of Counting:**

**1) Rule of Addition:**If a first task is performed in x ways and second task is performed in y ways, then either of the two operations can be performed in (x + y) ways

**2) Rule of Multiplication:**If a first task is performed in x ways and second task is performed in y ways, then both of the two operations can be performed in (x × y) ways

**Suppose n different cakes are done in a**

_{ 1 }, a_{ 2 }, a_{ 3 }, … a_{ n }different ways respectively, independent of each other then:1) Any one of them can be done in a

_{ 1}+ a

_{ 2 }+ a

_{ 3 }+ … + a

_{ n }ways. (a

_{ 1 }way or a

_{ 2 }way or a

_{ 3 }+ … + a

_{ n }way)

2) All of them can be done in a

_{1}× a

_{2}× a

_{3}× … × a

_{n }ways (a

_{1}way and a

_{2}way and a

_{3}+ … + a

_{n }way)

**All about Permutation:**

**1) Condition 1:**Number of permutations of n things, taken r at a time is given as follows:

^{n}P_{r} = n (n – 1) (n – 2) (n – 3)……. (n – r + 1) = | n! |

(n – r)! |

**2) Condition 2:**If there are N balls and out of these B

_{1}balls are alike, B

_{2}balls are alike , B

_{3}balls are alike and so on and B

_{r}are alike of r

^{th}kind, such that (B

_{1}balls + B

_{2}balls + B

_{3}balls ----- B

_{r}balls) = N balls.

In such condition,

Number of permutation of these N balls = | N! |

(B_{1})! × (B_{2})! × (B_{3})! × - - - - - (B_{r})! |

**3) Condition 3:**If number of permutations of n objects are all taken at a time,

then, ^{n}P_{r} = | n! | = n! |

0! |

**Important points to remember:**

**1)**If N different objects are to be arranged, then they can be arranged in N! ways.

**2)**N number of objects can be arranged around a circle in (N – 1)! ways.

**3)**Sometimes we have to solve problems on permutation considering the condition of

**Repetition**

**Repetition:**This condition is not used unless specified. (Remember)

Number of permutation of

**N objects taken r at a time**when each selected object can be repeated any number of times is given as:

**Number of permutations = n**

^{r}**4) Restricted Permutation:**The number of permutations of n objects taken r at a time in which if k particular objects are:

a) Never included:

^{ (n – k)}P

_{r}---- (k are the number of objects not included)

b) Always included:

^{ (n – k) }C

_{r–k x r! }---- (k is the number of objects always included)

**All about combinations:**

**1)**Number of combinations of n objects, taken r at a time is given as follows:

^{n}C_{r} = | n! | = | n(n – 1) (n – 2)…….to r factors |

(r!)(n – r)! | r! |

**This example will surely clear the concept!**

**Hint:**In the example discussed below, the confusion related to addition and multiplication of terms will also be cleared.

**Example:**Suppose there are 12 boys and 8 girls, and we have to select 5 volunteers for a particular task. So we have to find the number of possible selections we can make.

Total students are (12 + 8) = 20 and we have to select 5 volunteers.

The total number of selections can be made _{n}C^{r} ways = _{20}C^{5} | n! |

(r!)(n – r)! |

The question may be asked in different ways. 2 different conditions are specified below:

1) Out of 5 volunteers, 3 boys and 2 girls must be present.

2) Boys should be in majority.

Condition 1: 3 boys 2 girls are needed | Condition 2: If boys are in majority |
---|---|

Out of 12 boys 3 are selected and out of 8 girls 2 are selected. (Boys) ^{12}C_{3} and (Girls) ^{8}C_{2}Both girls and boys are needed, hence they are multiplied. ^{12}C_{3} × ^{8}C_{2} | 1) If only boys are selected as volunteers: ^{12}C_{5} 2) 4 boys and 3 girls: ^{12}C_{4} × ^{8}C_{1}3) 3 boys and 2 girls: ^{12}C_{3} × ^{8}C_{2}These will be the 3 possibilities, where boys are in majority. ( ^{12}C_{5} ) or (^{12}C_{4} × ^{8}C_{1}) or (^{12}C_{3} × ^{8}C_{2})( ^{12}C_{5}) + (^{12}C_{4} × ^{8}C_{1}) + (^{12}C_{3} × ^{8}C_{2}) |

**Important Formulae:**

**1)**

^{n}C

_{n}=

^{n}C

_{0}= 1

**2)**

^{n}C

_{n – 1}=

^{n}C

_{1}= n

**3)**

^{n}C

_{r}=

^{n}C

_{n – r}

**4)**0! = 1

**5)**n! = n (n – 1)!

6) ^{n}P_{r} = | n! |

(n – r)! |

7) ^{n}C_{r} = | nPr! |

r! |

**Types of Questions**

Type 1: Permutation

**Q 1.**Find the value of

^{50}P

_{2}

a. 4500

b. 3260

c. 2450

d. 1470

View solution

Correct Option: (c)**Hint: **

^{n}P_{r} = | n! |

(n – r)! |

50! | = | 50 × 49 × 48 | = 50 × 49 = 2450 |

(50 – 2)! | 48! |

**Value of**

^{50}P_{2}= 2450**Q 2.**How many words can be formed by using letters of the word ‘DELHI’?

a. 50

b. 72

c. 85

d. 120

View solution

Correct Option: (d)**Hint: **

The word ‘DELHI’ contains 5 letters

Therefore, required number of words = ^{5}P_{5} = 5! = (5 × 4 × 3 × 2 × 1) = 120

120 words can be formed by using letters of the word ‘DELHI’

**Q 3.**Find the number of ways the letters of the word ‘RUBBER can be arranged?

a. 450

b. 362

c. 250

d. 180

View solution

Correct Option: (d)**Hint: **

The word ‘RUBBER’ contains 6 letters: 2R, 2B, 1 U, 1 E

Therefore,

The required Number of ways: | N! |

(2R!) × (2B!) × (1U!) × (1E!) |

= | 6! |

(2 × 1) × (2 × 1) × (1) × × (1) |

= | 6 × 5 × 4 × 3 × 2 × 1 |

4 |

= 180

**Q 4.**Find in how many different ways, the letters of the word ‘LEADING’ can be arranged in such way that the vowels always come together?

a. 548

b. 426

c. 720

d. 790

View solution

Correct Option: (c)

The letters in the word ‘LEADING’ are 7.

Number of vowels = EAI

Number of consonants: LDNG**Condition: **Vowels always come together. Therefore, let’s consider the vowels as one group ‘EAI’

Hence, vowels ‘EAI’ and consonants LDNG together form 5 letters.

Therefore, these 5 letters can be arranged in 5! Ways = 5! = 5 × 4 × 3 × 2 = 120 ways

Letters in group of vowels ‘EAI’ can also be interchanged. So, these vowels can be arranged in 3! Ways = 6 ways.

All together, required number of ways = 120 x 6 = 720 ways