Dictionary Rank of a Word | Permutations & Combinations

 PERMUTATIONS & COMBINATIONS Rank of the word or Dictionary order of the English words like COMPUTER, COLLEGE, SUCCESS, SOCCER, RAIN, FATHER, etc. Dictionary Rank of a Word Method of finding the Rank (Dictionary Order) of the word  “R A I N” Given word: R A I N Total letters = 4 Letters in alphabetical order: A, I, N, R No. of words formed starting with A = 3! = 6 No. of words formed starting with I = 3! = 6 No. of words formed starting with N = 3! = 6 After N there is R which is required R ----- Required A ---- Required I ---- Required N ---- Required RAIN ----- 1 word   RANK OF THE WORD “R A I N” A….. = 3! = 6 I……. = 3! = 6 N….. = 3! = 6 R…A…I…N = 1 word 6 6 6 1 TOTAL 19 Rank of “R A I N” is 19 Method of finding the Rank (Dictionary Order) of the word  “F A T H E R” Given word is :  "F A T H E R" In alphabetical order: A, E, F, H, R, T Words beginni

Relation and Functions Class XII Chapter 1


Class 12  chapter 1

Relations and functions

Basic concepts of topic relations and functions class XII chapter 1 of mathematics. Equivalence relations, different types of functions, composition and inverse of functions.
Introduction
In class 11 we have studied about Cartesian product of two sets, relations, functions, domain, range and co-domains. Now in this chapter we have studied about the different types of relations, different types of functions, composition of functions and invertible functions.


Relations
A relation R from set A to set B is the subset of Cartesian product A x B.
In Cartesian product A x B number of relations equal to the number of subsets.

Domain
Set of all first elements of the ordered pairs in the relation R is called its domain.

Range
Set of all second elements of the ordered pairs in the relation R is called its range.
In any ordered pair second element is also called the image of first element.



Types of relations
i) Empty Relation:
A relation R in set A is called empty relation, if no element of A is related to any element of A, i.e., R = φ  ⊂ A x A
Example: Let A = {1, 2, 3} and R be a relation on the set A defined as R = {(a, b) : a + b = 10; a, b ∈ A}
Here R is empty relation. R = Φ

ii) Identity Relation
A relation R in set A is called identity relation, if every element of A is related to itself, i.e., R = {a, a)  ⊂ A x A

A relation R is an identity relation in set A if for all a ∈ A, (a, a) ∈ R.

iii) Universal Relation :
A relation R in a set A is called universal relation, if each element of A is related to every element of A, i.e., R = A x A

iv) Reflexive relation
A relation R in a set A is called reflexive, if (a, a) ∈ R,  for every a ∈ A

v) Symmetric Relation
A relation R in a set A is called symmetric, if (a, b) ∈R implies that (b, a) ∈ R, for all a, b ∈ A.

vi) Transitive Relation
A relation R in a set A is called transitive, if (a, b) ∈R and (b, c) ∈R, implies that (a, c) ∈ R, for all a, b, c ∈ A

vii) Equivalence relation:
A relation R in set A is said to be equivalence relation if it is reflexive, symmetric  and transitive relation.

Viii) Equivalence Relations:
Let R be an equivalence relation on a set A and let a ∈ A. Then we define equivalence class of a as 
[a] ={b ∈ A : b is related to a} = {b ∈ A : (b, a) ∈ R}

Note : Both empty relation and universal relations are sometimes called equivalence relation of trivial relation.





Functions 
One-One functions (or injective functions):
A function f : A →B is said to  be one –one or injective function if different elements of set A has different images in set B
Mathematically: 
Case 1: For all a, b ∈ A, we have: If  f(a) = f(b) a =  b
Case 2 :  For a, b ∈ A, such that : If ≠ b ⇒ f(a) ≠ f(b)

Relations & Functions

In the above figure different elements of set A have different images in set B. So this function is called one-one function.

Set A has 3 elements and the set B has 4 elements. Then the number of injective functions that can be defined from set A to set B is  = 4P3 = 4! = 24.
Generalization
Set A has m elements and the set B has n elements (n > m). Then the number of injective functions that can be defined from set A to set B is  =  nPm  = n! 

If n < m then Then the number of injective functions that can be defined from set A to set B is = 0

If a function has m elements and is defined from A to A then number of injective relations from A to A = n!

Many One functions:
If two or more elements of set A have same image in set B then it is called many one function.
If a function is not one-one then it is called many one.
Relation and Functions


Here two different elements a and b of set A has same image in set B. So it is not an one-one function. It is only a many one function.

Onto function ( or surjective function)
A function f : A →B is said to be onto (or surjective) function, if every element of set B is the image of some element in set A.

Let y ∈ B, then for every  element  y ∈ B,  ∃ (there exist) an element   x ∈ A such that  f(x) = y
Relations & Functions

In the above figure every element of set B is the image of some element in set A so it is a onto function. It is neither  many one nor   one-one.

If n(A) = p, then number of bijective functions from set A to A are p!
Because If n(A) = p and n(A) = n(B) then the number of bijective functions from A to B is p!

Other definitions of onto functions are as follows
A function f : A →B is said to be onto (or surjective) function, if every element of set B has pre - image in set A.

If range of the function is equal to the co-domain(or set B), then it is called onto function.
Exercise 1.2
Q No. 9
\[f:N\rightarrow N,\; \; f(n)=\left\{\begin{matrix} \frac{n+1}{2}, if\; \; n\; \; is\; \; odd\\ \frac{n}{2}, if\; \; n \; \; is\; \; even \end{matrix}\right.\]
One-One
\[Let\; (1,2)\epsilon N\; then\]\[f(1)=\frac{n+1}{2}=\frac{1+1}{2}=\frac{2}{2}=1\]\[f(2)=\frac{n}{2}=\frac{2}{2}=1\]\[f(1)=f(2)\: but\: \: 1\neq 2\]⇒ f is not one one function.
Onto
Case 1: If n = odd = 2q+1=4q+1\[f(n)=\frac{n+1}{2}\]\[f(4q+1)=\frac{4q+1+1}{2}=\frac{4q+2}{2}=2q+1\]⇒ 2q + 1 is also an odd function ∈ N
Case 2: If n = even = 2q = 4q, \[f(n)=\frac{n}{2}=\frac{4q}{2}=2q\]⇒ 2q  is also an even function ∈ N
From case 1 and case 2 we conclude that :
*For all odd numbers in the domain we have odd numbers in the co-domain .
*For all even numbers in the domain we have even numbers in the co-domain .
⇒ Rf = Odd natural number and Even natural numbers.
⇒ Rf = All Natural Numbers = Co-domain
⇒ f(x) is an onto function.
Now f(x) is an onto function but not one-one.
So f(x) is not a bijective function.

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Composition of two functions
Let f : A →B and  g : B →C be two functions. Then the composition of f and g, denoted by gof, is defined as the function gof : A →C given by
gof(x) = g[f(x)],  ∀ (for all)  x ∈ A

Composition of two functions exists only if co-domain of first function is equal to the domain of second function
Relations & Functions

In the above figure function f is defined on the element of set A (i.e.  x), function g is defined on the elements of set B (i.e f(x) ) so composition from set A to set C becomes gof or g[f(x)]

Bijective functions:
The functions which are one-one and onto both are called bijective functions.

Invertible functions and existence of inverse
If a function f is one-one and onto then it is an invertible function.

If a function is an invertible function then inverse of f exist and is denoted by f-1.

If f is invertible the f must be one-one and onto, conversely if f is one-one and onto  then f must be invertible. If f is invertible then f-1 exists.

Let f(x) = y is an invertible function then f-1 exists and we can write f-1(y) = x

Example : f(x) = { (a, 1), (b, 2), (c, 3), (d, 4)}, this can be written as
f(a) = 1, f(b) = 2,  f(c) = 3,  f(d) = 4

If f is invertible then we can write
f-1(1) = a,  f-1(2) = b,  f-1(3) = c,  f-1(4) = d
If f and g are two invertible functions then their composition gof is also invertible.

Give an example of a relation which is
i) Symmetric but neither reflexive nor transitive
Ans  R = { (1,2), (2,1)}
ii) Transitive but neither reflexive nor symmetric
Ans:  R = { (1,1), (2,2), (1,2), (2,1), (3,4)}
iii) Reflexive and symmetric but not transitive.
Ans: R = {(1,1), (2,2), (3,3), (1,2), (2,1), (2,3), (3,2)}
iv) Reflexive and transitive but not symmetric.
Ans: R = {(1,1), (2,2), (1,2), (2,1), (2,3), (3,3)}
v) Symmetric and transitive but not reflexive

{(5,6), (6,5),(5,5) }



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