Notation and Set Theory

Definition: Sets and Operations on Sets

• A set is a collection of objects chosen from some universe. The universe is usually understood from the context. Sets are denoted by capital, bold letters or curly brackets.


• A B: A is a subset of B means that every element in A is also contained in B.


• A B: A union B is the set of all elements that are either in A or in B or in both.


• A B: A intersection B is the set of all elements that are in both sets A and B.


• A B: A minus B are all elements from A that are not in B.


• comp(A): The complement of A consists of all elements that are not in A.


• Two sets are disjoint if A B = 0 (the empty set)


• Two sets A and B are equal if A B and B A

• N = {1, 2, 3, 4, ... } = natural numbers (sometimes 0 is considered part of the natural numbers as well)


• Z = {... -3, -2, -1, 0, 1, 2, 3, ... } = integers


• Q = {p / q : p, q Z} (read as “all number p / q, such that p and q are elements of Z) = rational numbers


• R = real numbers


• 0 = empty set (the set that contains no elements)

Examples:

• Define the following sets: E = { x: x = 2n for n N}, O = { x: x = 2n + 1 for n N}, A = { x R : -4 < x < 3}, B = { x R : -1 < x < 7}, and I = { x R: x² = -2}. Then:


• 
  
  
  
  1.  What, in words, are the sets E, O, and I ?
  
  
  2.  Find A B, A B, A  B, comp(A).
  
  
  3.  Find O E, O I, comp(I).
  
  


• 
  
  
  
  • The set E is the set of all even integers.
  
  
  • The set O is the set of all odd integers.
  
  
  • The set I consists of all reals numbers whose square equals -2. Since there are no such real numbers, the set I is the empty set
  
  
  • Sets can be combined using the above operations much like adding and multiplying numbers. Familiar laws such as associative, commutative, and distributive laws will be true for sets as well. As an example, the next result will illustrate the distributive law; other laws are left as exercises.
  
  
  • 
    
    
    

    Proposition: Distributive Law for Sets

    
    
    
    
    • A (B C) = (A B) (A C)
    
    
    • A (B C) = (A B) (A C)
    
    
    • 
      Many results in set theory can be illustrated using Venn diagram, as in the above proof. However, such diagrams do not represent mathematically rigorous proofs. Nonetheless, before an actual proof is developed, it is first necessary to form a mental picture of the assumptions, conclusions, and implications of a theorem. For this process a Venn diagram can be very helpful. You can practice Venn diagrams by using them for some of the true/false statements in the exercises.
      There are many other theorems dealing with operation on sets. One that is particularly interesting is the theorem about de Morgan's Laws, because it deals with any number of sets (even infinitely many). Drawing a Venn diagram in such a situation would be impossible, but a mathematical proof can easily deal with this situation:
    
    
    • 
      
      
      
      

      Theorem: de Morgan’s Laws

      
      

      
      
      •  i.e. the complement of the intersection of any number of sets equals the union of their complements.
      
      
      •  i.e. the complement of the union of any number of sets equals the intersection of their complements.
      
      

      
      
      

      Relations and Functions

      
      

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      After introducing some of the basic elements of set theory (sets), we will move on to the second most elementary concept, the concept of relations and functions.
      
      

      Definition: Relation

      
      

      
      
      • Let A and B be two sets. A relation between A and B is a collection of ordered pairs (a, b) such that a A and b B. Often we use the notation a ~ b to indicated that a and b are related, rather then the order pair notation (a, b).
      
      

      
      Note that this does not mean that each element from A needs to be associated with one (or more) elements from B. It is sufficient if some associations between elements of A and B are defined. In contrast, there is the definition of a function:
      
      
      

      Definition: Function, Domain, and Range

      
      

      
      
      • Let A and B be two sets. A function f from A to B is a relation between A and B such that for each a A there is one and only one associated b B. The set A is called the domain of the function, Bis called its range.
      
      
      • Often a function is denoted as y = f(x) or simply f(x), indicating the relation { (x, f(x)) }.
      
      

      
      
      

      Examples:

      
      

      
      
      • Let A = {1, 2, 3, 4}, B = {14, 7, 234}, C = {a, b, c}, and R = real numbers. Define the following relations:
        
        
        
        1. r is the relation between A and B that associates the pairs 1 ~ 234, 2 ~ 7, 3 ~ 14, 4 ~ 234, 2 ~ 234
        
        
        2. f is the relation between A and C that relates the pairs {(1,c), (2,b), (3,a), (4,b)}
        
        
        3. g is the relation between A and C consisting of the associations {(1,a), (2,a), (3,a)}
        
        
        4. h is the relation between R and itself consisting of pairs {(x,sin(x))}
        
        
      
      
      •  Which of those relations are functions ?
      
      

      
      The outcomes of a function (i.e. the elements from the range associated to elements in the domain) do not only depend on the rule of the function (such as x is associated with sin(x)) but also on the domain of the function. Therefore, we need to specify those outcomes that are possible for a given rule and a given domain:
      
      
      

      Definition: Image and Preimage

      
      

      
      
      • Let A and B be two sets and f a function from A to B. Then the image of f is defined as
        
        
        
        • imag(f) = {b B : there is an a A with f(a) = b}.
        
        
      
      
      • Let A and B be two sets and f a function from A to B. If C is a subset of the range B then the preimage, or inverse image, of C under the function f is the set defined as
        
        
        
        • f ^-1(C) = {x A : f(x) C }
        
        
      
      

      
      As an example, consider the following functions:
      
      

      Example:

      
      

      
      
      • Let f(x) = 0 if x is rational and f(x) = 1 if x is irrational. This function is called Dirichlet’s Function. The range for f is R.
        
        
        
        •  Find the image of the domain of the Dirichlet Function when:
          
          
          
          1. the domain of f is Q
          
          
          2. the domain of f is R
          
          
          3. the domain of f is [0, 1] (the closed interval between 0 and 1)
          
          
        
        
        •  What is the preimage of R ? What is the preimage of [-1/2, 1/2] ?
        
        
      
      
      • Let f(x) = x², with domain and range being R. Then use the graph of the function to determine:
        
        
        
        1.  What is the image of [0, 2] and the preimage of [1, 4] ?
        
        
        2.  Find the image and the preimage of [-2, 2].
        
        
      
      

      
      Functions can be classified into three groups: those for which every element in the image has one preimage, those for which the range is the same as the image, and those which have both of these properties. Accordingly, we make the following definitions:
      
      

      Definition: One-one, Onto, Bijection

      
      

      
      
      • A function f from A to B is called one to one (or one- one) if whenever f(a) = f(b) then a = b. Such functions are also called injections.
      
      
      • A function f from A to B is called onto if for all b in B there is an a in A such that f(a) = b. Such functions are also called surjections.
      
      
      • A function f from A to B is called a bijection if it is one to one and onto, i.e. bijections are functions that are injective and surjective.
      
      

      
      

      Examples:

      
      

      
      
      • 
        
         Which of the following functions are one-one, onto, or bijections ? The domain for all functions is R.
        
        
        
        1. f(x) = 2x + 5
        
        
        2. g(x) = arctan(x)
        
        
        3. g(x) = sin(x)
        
        
        4. h(x) = 2x³ + 5x² - 7x + 6
        
        
      
      

      
      

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      Problems:

      
      

      1. Define the following sets: E = { x: x = 2n for n N}, O = {x : x = 2n + 1 for n N}, A = {x R : -4 < x < 3}, B = {x R : -1 < x < 7}, and I = {x R: x * x = -2}. Then:
      
      

      
      
      
      
      1. What, in words, are the sets E, O, and I ?
      
      
      2. Find A B, A B, A  B, comp(A).
      
      
      3. Find O E, O I, comp(I).
      
      
      
      
      1. 2. Let A, B, and C be any sets. Prove the following statements:
         
         
         
         
         
         
         1. C (A B) = (C A) (C B)
         
         
         2. C (A B) = (C A) (C B)
         
         
         
         
         
         
         
         
         1. 3. Prove that when two even integers are multiplied, the result is an even integer, and when two odd integers are multiplied, the result is an odd integer.
            
            
            
            
            

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            4. Prove or disprove: if the square of a number is an odd integer, then the original number must also be an odd integer. (Try a proof by contradiction)
            
            
            
            
            

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            5. Prove that there is no greatest positive integer. Also prove that there is no smallest positive rational number.
            
            

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            6. Prove or disprove: every prime number greater than 2 is odd.
            
            

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            7. Euclid's Theorem states that there is no largest prime. A proof by contradiction would start out by assuming that the statement is false, i.e. there is a largest prime. The advantage now is that if there was a largest prime, there would be only finitely many primes. This seems easier to handle than the original statement which implies the existence of infinitely many primes. Finish the proof.
            
            
            
            8. Let A = {1, 2, 3, 4}, B = {14, 7, 234}, C = {a, b, c}, and R = real numbers. Which of the following relations are functions ?
            
            
            
            
            
            
            1. r is the relation between A and B that associates the pairs 1 ~ 234, 2 ~ 7, 3 ~ 14, 4 ~ 234, 2 ~ 234
            
            
            2. f is the relation between A and C that relates the pairs {(1,c), (2,b), (3,a), (4,b)}
            
            
            3. g is the relation between A and C consisting of the associations {(1,a), (2,a), (3,a)}
            
            
            4. h is the relation between R and itself consisting of pairs {(x,sin(x))}
            
            
            
            
            
            
            
            9. Give an example of a relation r on a set A such that:
            
            
            
            
            
            
            1. the relation r is reflexive and symmetric but not transitive
            
            
            2. the relation r is symmetric and transitive but not reflexive
            
            
            3. the relation r is transitive but neither reflexive nor symmetric.
            
            
            
            
            
            

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            10. Let f(x) = 0 if x is rational and f(x) = 1 if x is irrational. This function is called Dirichlet’s Function. The range for f is R.
            
            
            
            
            
            
            1. What is the image of the function if the domain is Q
            
            
            2. What is the image of the function if the domain is R
            
            
            3. What is the image of the function if the domain is [0, 1] (the closed interval between 0 and 1)
            
            
            4. What is the preimage of R ? What is the preimage of [-1/2, 1/2] ?
            
            
            
            
            
            
            
            
            
            

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            11. Answer the above questions for the function f(x) = 0 if x is rational and f(x) = x if x is irrational.
            
            

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            12. Let f(x) = 1 - , with domain and range being R. Then use the graph of the function to determine:
            
            
            
            
            
            
            1. What is the image of [0, 2] and the preimage of [1, 4] ?
            
            
            2. Find the image and the preimage of [-2, 2].
            
            
            
            
            
            
            
            13. If the graph of a function is known, how can you decide from that graph whether a function is one-to-one (injective) or onto (surjective) ?
            
            
            
            14. Which of the following functions are one-one, onto, or bijections ? The domain for all functions is R.
            
            
            
            
            
            
            1. 
            
            
            2. 
            
            
            3. 
            
            
            4. 
            
            
            
            
            
            
            
            15. Show that the range of a function is equal to the image of the domain of the function if and only if the function is surjective.
            
            

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            16. Show that every bijection has an inverse function.
            
            

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            17. Show that if f is a bijection from A to B then is a bijection mapping B onto A. Moreover, show that the inverse of is the original function f.
            
            

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            18. Let A = {1, 2, 3, 4} and B = {a, b, c}. Which of the following relations is an equivalence relation ?
            
            
            
            
            
            
            1. r : { (a,a), (b,b), (a,b), (b,a) }
            
            
            2. s : 1 ~ 1, 2 ~ 2, 3 ~ 3, 4 ~ 4, 1 ~ 4, 4 ~ 1, 2 ~ 4, 4 ~ 2
            
            
            
            
            
            
            
            19. If S = {a, b, c}, then list all possible equivalence relations on the set S.
            
            

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            20. Consider the set Z of all integers. Define a relation r by saying that x and y are related if their difference y - x is divisible by 3. Then
            
            
            
            
            
            
            1. Check that this relation is an equivalence relation
            
            
            2. Find the two equivalence classes, and name them appropriately.
            
            
            3. How would you add these equivalence classes, if at all ?
            
            
            
            
            
            
            
            21. Define a relation r on R by saying that two real numbers x and y are related if there exists an integer n such that | x - n | < 1/2 and | y - n | < 1/2. Show that this relation is an equivalence relation and describe the equivalence classes.
            
            
            
            22. Let S be the collection of all polynomials with real coefficients. We say that two polynomials p(x) and q(x) are related if the number 0 is a root of the polynomial p(x) - q(x). Is this an equivalence relation on S ?
            
            

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            23. The Rational Numbers together with addition and multiplication can be defined as follows:
            
            
            
            
            1. Let A be the set N x N and define a relation r on N x N by saying that (a, b) is related to (a’, b’) if a * b’ = a’ * b. Show that this relation is an equivalence relation.
            
            
            2. If [(a, b)] and [(a’, b’)] denotes the equivalence classes containing (a, b) and (a’, b’), respectively, then show that the following operations are well-defined:
               
               
               
               • [(a, b)] + [(a', b')] = [(a * b' + a’ * b, b * b')]
               
               
               • [(a, b)] * [(a’, b’)] = [(a * a’, b * b’)]
               
               
            
            
            3. Give examples to indicate that the resulting set of all equivalence classes has all of the familiar properties of the rational numbers (and hence can serve to define the rationals based only on the natural numbers).
            
            
         
         
      
      

      
      

      
      
      
      
      
      
      
    
    
  
  

 A block of mass 15 kg is placed on a trolley.THe coefficient of friction b/wblock & trolley is 0.18.The trolley acclerates frm rest with 0.5 m/s² for 20 seconds and then moves with uniform velocity.then:

(a) to a stationary observer on ground,fricton of 7.5 N is acting on the block,so that the block & the trolley have the same accleration.

(b)to a stationary observer on ground,when trolley moves with uniform velocity,no force of friction acts on the block

(c)to an observer on the trolley,the block will be at rest

(d)to an observer on the trolley,no friction will act on the block when the trolley moves with uniform velocity

all the optins are correct..............

but the concept of the (d) is not clear for me

could anyone help me

 Permutation and Combination

<<(restricted combination) previous | next (circular permutation)>>

Permutation : Permutation means arrangement of things. The word arrangement is used, if the order of things is considered.
Combination: Combination means selection of things. The word selection is used, when the order of things has no importance.

Example:     Suppose we have to form a number of consisting of three digits using the digits 1,2,3,4, To form this number the digits have to be arranged. Different numbers will get formed depending upon the order in which we arrange the digits. This is an example of Permutation.

Now suppose that we have to make a team of 11 players out of 20 players, This is an example of combination, because the order of players in the team will not result in a change in the team. No matter in which order we list out the players the team will remain the same! For a different team to be formed at least one player will have to be changed.

Now let us look at two fundamental principles of counting:

Addition rule : If an experiment can be performed in ‘n’ ways, & another experiment can be performed in ‘m’ ways then either of the two experiments can be performed in (m+n) ways. This rule can be extended to any finite number of experiments.

Example:       Suppose there are 3 doors in a room, 2 on one side and 1 on other side. A man want to go out from the room. Obviously he has ‘3’ options for it. He can come out by door ‘A’ or door ‘B’ or door ’C’.

Multiplication Rule : If a work can be done in m ways, another work can be done in ‘n’ ways, then both of the operations can be performed in m x n ways. It can be extended to any finite number of operations.

Example.:      Suppose a man wants to cross-out a room, which has 2 doors on one side and 1 door on other site. He has  2 x 1  = 2 ways for it.

Factorial n : The product of first ‘n’ natural numbers is denoted by n!.
            n!   = n(n-1) (n-2) ………………..3.2.1.

            Ex.       5! = 5 x 4 x 3 x 2 x 1 =120

            Note       0!     =  1

            Proof   n! =n, (n-1)!

            Or           (n-1)! = [n x (n-1)!]/n = n! /n                     

            Putting n = 1, we have

            O!  = 1!/1

            or  0 = 1      

Permutation

Number of permutations of ‘n’ different things taken ‘r’ at a time is given by:-

ⁿP_r       =            n!/(n-r)!

Proof:     Say we have ‘n’ different things a₁, a₂……, a_n.

Clearly the first place can be filled up in ‘n’ ways. Number of things left after filling-up the first place = n-1

So the second-place can be filled-up in (n-1) ways. Now number of things left after filling-up the first and second places = n - 2

Now the third place can be filled-up in (n-2) ways.

Thus number of ways of filling-up first-place = n

Number of ways of filling-up second-place = n-1

Number of ways of filling-up third-place = n-2

Number of ways of filling-up r-th place = n – (r-1) = n-r+1

By multiplication – rule of counting, total no. of ways of filling up, first, second --  rth-place together :-

n (n-1) (n-2) ------------ (n-r+1)

Hence:

ⁿP_r       = n (n-1)(n-2) --------------(n-r+1)

= [n(n-1)(n-2)----------(n-r+1)] [(n-r)(n-r-1)-----3.2.1.] / [(n-r)(n-r-1)] ----3.2.1

ⁿP_r = n!/(n-r)!

Number of permutations of ‘n’ different things taken all at a time is given by:-

ⁿP_n               =         n!

Proof   : 

Now we have ‘n’ objects, and n-places.

Number of ways of filling-up first-place  = n

Number of ways of filling-up second-place = n-1

Number of ways of filling-up third-place  = n-2

Number of ways of filling-up r-th place, i.e. last place =1

Number of ways of filling-up first, second, --- n th place

= n (n-1) (n-2) ------ 2.1.

ⁿP_n =  n!

Concept. 

We have   ⁿP_r  =     n!/n-r

Putting r = n, we have :-

ⁿP_r  =   n! / (n-r)

But   ⁿP_n  =  n!

Clearly it is possible, only when  n!  = 1

Hence it is proof that     0! = 1

Note : Factorial of negative-number is not defined. The expression  –3! has no meaning.

Examples

Q. How many different signals can be made by 5 flags from 8-flags of different colours?

Ans.    Number of ways taking 5 flags out of 8-flage  = ⁸P₅

=   8!/(8-5)!       

=  8 x 7 x 6 x 5 x 4 = 6720

Q. How many words can be made by using the letters of the word “SIMPLETON” taken all at a time?

Ans.   There are ‘9’ different letters of the word “SIMPLETON”

Number of Permutations taking all the letters at a time  = ⁹P₉

=  9!    = 362880.

Number of permutations of n-thing, taken all at a time, in which ‘P’ are of one type, ‘g’ of them are of second-type, ‘r’ of them are of third-type, and rest are all different is given by :-

  n!/p! x q! x r!                                                                                               

Example: In how many ways can the letters of the word “Pre-University” be arranged?

13!/2! X 2! X 2!

Number of permutations of n-things, taken ‘r’ at a time when each thing can be repeated r-times is given by = n^r.

Proof.   

Number of ways of filling-up first –place   = n

Since repetition is allowed, so

Number of ways of filling-up second-place  = n

Number of ways of filling-up third-place             

Number of ways of filling-up r-th place  = n

Hence total number of ways in which first, second ----r th, places can be filled-up

 =  n x n x n ------------- r factors.

=   n^r

Example:      A child has 3 pocket and 4 coins. In how many ways can he put the coins in his pocket.

Ans.    First coin can be put in 3 ways, similarly second, third and forth coins also can be put in 3 ways.

So total number of ways = 3 x 3 x 3 x 3   = 3⁴   = 81

Circular permutations

There are two cases of circular-permutations:-

(a)       If clockwise and anti clock-wise orders are different, then total number of circular-permutations is given by (n-1)!

(b)       If clock-wise and anti-clock-wise orders are taken as not different, then total number of circular-permutations is given by  (n-1)!/2!

Proof(a):         

(a)       Let’s consider that 4 persons A,B,C, and D are sitting around a round table

Shifting A, B, C, D, one position in anticlock-wise direction, we get the following agreements:-

Thus, we use that if 4 persons are sitting at a round table, then they can be shifted four times, but these four arrangements will be the same, because the sequence of A, B, C, D, is same. But if A, B, C, D, are sitting in a row, and they are shifted, then the four linear-arrangement will be different.

Hence if we have ‘4’ things, then for each circular-arrangement number of linear-arrangements =4

Similarly, if we have ‘n’ things, then for each circular – agreement, number of linear – arrangement = n.

Let the total circular arrangement  = p

Total number of linear–arrangements = n.p

Total number of linear–arrangements 

= n. (number of circular-arrangements)

Or Number of circular-arrangements = 1 (number of linear arrangements)

 n  = 1( n!)/n

circular permutation = (n-1)!

Proof (b)   When clock-wise and anti-clock wise arrangements are not different, then observation can be made from both sides, and this will be the same. Here two permutations will be counted as one. So total permutations will be half, hence in this case.

Circular–permutations =  (n-1)!/2

Note:   Number of circular-permutations of ‘n’ different things taken ‘r’ at a time:-

(a)  If clock-wise and anti-clockwise orders are taken as different, then total number of circular-permutations  =   ⁿP_r/r

(b) If clock-wise and anti-clockwise orders are taken as not different, then total number of circular – permutation =      ⁿP_r/2r  

Example: How many necklace of 12 beads each can be made from 18 beads of different colours?

Ans.    Here clock-wise and anti-clockwise arrangement s are same.

Hence total number of circular–permutations:       ¹⁸P₁₂/2x12

 =   18!/(6    x   24)

Restricted – Permutations

(a)   Number of permutations of ‘n’ things, taken ‘r’ at a time, when a particular thing is to be always included in each arrangement

= r ^n-1 P_r-1

(b) Number of permutations of ‘n’ things, taken ‘r’ at a time, when a particular thing is fixed: = ^n-1 P_r-1

(c) Number of permutations of ‘n’ things, taken ‘r’ at a time, when a particular thing is never taken: = ^n-1 P_r.

(d) Number of permutations of ‘n’ things, taken ‘r’ at a time, when ‘m’ specified things always come together = m!  x (  n-m+1) !

(e) Number of permutations of ‘n’ things, taken all at a time, when ‘m’ specified things always come together = n ! - [ m! x   (n-m+1)! ]

Example:   How many words can be formed with the letters of the word ‘OMEGA’ when:

(i)       ‘O’ and ‘A’ occupying end places.

(ii)       ‘E’ being always in the middle

(iii)       Vowels occupying odd-places

(iv)        Vowels being never together.

Ans.

(i)    When ‘O’ and ‘A’ occupying end-places

  => M.E.G. (OA)

  Here (OA) are fixed, hence M, E, G can be arranged in  3! ways

 But (O,A) can be arranged themselves is 2! ways.

=> Total number of words =  3!   x  2! = 12 ways.

(ii)  When ‘E’ is fixed in the middle

=>    O.M.(E), G.A.

Hence four-letter O.M.G.A. can be arranged in  4!   i.e 24 ways.

(iii)   Three vowels (O,E,A,) can be arranged in the odd-places (1^st, 3^rd and 5^th)    =  3!    ways.

And two consonants (M,G,) can be arranged in the even-place               (2^nd, 4^th) =   2 !   ways

=> Total number of ways= 3! x 2! = 12 ways.

(iv)  Total number of words   =   5!   =    120!

 If all the vowels come together, then we have: (O.E.A.), M,G

 These can be arranged in    3!    ways.

 But (O,E.A.) can be arranged themselves in   3! ways.

 => Number of ways, when vowels come-together  =    3!  x    3!  

= 36 ways

=> Number of ways, when vowels being never-together

= 120-36        =  84 ways.

Number of Combination of ‘n’ different things, taken ‘r’ at a time is given by:-

ⁿC_r=  n! / r ! x (n-r)!                                

Proof: Each combination consists of ‘r’ different things, which can be arranged among themselves in   r!  ways.

=> For one combination of ‘r’ different things, number of arrangements =    r!

For ⁿC_r combination number of arrangements:      r    ⁿC_r

=> Total number of permutations =    r!   ⁿC_r  ---------------(1) 

But number of permutation of ‘n’ different things, taken ‘r’ at a time

= ⁿP_r -------(2)

From (1) and (2) :

ⁿP_r  =      r!  .  ⁿC_r

or      n!/(n-r)!  =  r!   .  ⁿC_r          

or   ⁿC_r    =       n!/r!x(n-r)!

Note: ⁿC_r  =  ⁿC_n-r

or   ⁿC_r    = n!/r!x(n-r)!   and  ⁿC_n-r  =   n!/(n-r)!x(n-(n-r))!

 =  n!/(n-r)!xr!

Restricted – Combinations

(a)  Number of combinations of ‘n’ different things taken ‘r’ at a time, when ‘p’ particular things are always included = ^n-pC_r-p.

(b)  Number of combination of ‘n’ different things, taken ‘r’ at a time, when ‘p’ particular things are always to be excluded = ^n-pC_r

Example:      In how many ways can a cricket-eleven be chosen out of 15 players? if

(i)  A particular player is always chosen,

(ii)  A particular is never chosen.

Ans:   

(i)       A particular player is always chosen, it means that 10 players are selected out of the remaining 14 players.

=. Required number of ways =  ¹⁴C₁₀  = ¹⁴C₄

= 14!/4!x19!  = 1365                                              

(ii) A particular players is never chosen, it means that 11 players are selected out of 14 players.

 => Required number of ways =  ¹⁴C₁₁ 

 =   14!/11!x3!  = 364

(iii) Number of ways of selecting zero or more things from ‘n’ different things is given by:-   2ⁿ-1

Proof:  Number of ways of selecting one thing, out of n-things     = ⁿC₁

Number of selecting two things, out of n-things =ⁿC₂

Number of ways of selecting three things, out of n-things =ⁿC₃

Number of ways of selecting ‘n’ things out of ‘n’ things = ⁿC_n

=>Total number of ways of selecting one or more things out of n different things

= ⁿC₁ + ⁿC₂ + ⁿC₃ + ------------- + ⁿC_n

= (ⁿC₀ + ⁿC₁ + -----------------ⁿC_n)  - ⁿC₀

= 2ⁿ – 1                           [ ⁿC₀=1]

Example:  John has 8 friends. In how many ways can he invite one or more of them to dinner?

Ans.    John can select one or more than one of his 8 friends.

=> Required number of ways = 2⁸ – 1= 255.

(iv) Number of ways of selecting zero or more things from ‘n’ identical things is given by :- n+1

Example:   In how many ways, can zero or more letters be selected form the letters AAAAA?

Ans. Number of ways of :

         Selecting zero 'A's   =  1

         Selecting one 'A's   =   1

         Selecting two 'A's     = 1

         Selecting three 'A's  =  1

         Selecting four 'A's   =   1

         Selecting five 'A's    =  1

=> Required number of ways  = 6         [5+1]

(V)    Number of ways of selecting one or more things from ‘p’ identical things of one type ‘q’ identical things of another type, ‘r’ identical things of the third type and ‘n’ different things is given by :-

 (p+1) (q+1) (r+1)2ⁿ – 1

Example:    Find the number of different choices that can be made from 3 apples, 4 bananas and 5 mangoes, if at least one fruit is to be chosen.

Ans:

Number of ways of selecting apples = (3+1) = 4 ways.

Number of ways of selecting bananas = (4+1) = 5 ways.

Number of ways of selecting mangoes = (5+1) = 6 ways.

Total number of ways of selecting fruits = 4 x 5 x 6

But this includes, when no fruits i.e. zero fruits is selected

=> Number of ways of selecting at least one fruit = (4x5x6) -1  = 119

Note :- There was no fruit of a different type, hence here  n=o

 =>   2ⁿ  = 2⁰=1

(VI)   Number of ways of selecting ‘r’ things from ‘n’ identical things is ‘1’.

Example:   In how many ways 5 balls can be selected from ‘12’ identical red balls?

Ans. The balls are identical, total number of ways of selecting 5 balls  = 1.

Example: How many numbers of four digits can be formed with digits 1, 2, 3, 4 and 5?

Ans. Here n = 5                     [Number of digits]

And   r = 4                     [ Number of places to be filled-up]

Required number is   ⁵P₄ = 5!/1!  = 5 x 4 x 3 x 2 x 1

Restricted Permutations

(a)   Number of permutations of ‘n’ things, taken ‘r’ at a time, when a particular thing is to be always included in each arrangement

= r ^n-1 P_r-1

(b) Number of permutations of ‘n’ things, taken ‘r’ at a time, when a particular thing is fixed: = ^n-1 P_r-1

(c) Number of permutations of ‘n’ things, taken ‘r’ at a time, when a particular thing is never taken: = ^n-1 P_r.

(d) Number of permutations of ‘n’ things, taken ‘r’ at a time, when ‘m’ specified things always come together = m!  x (  n-m+1) !

(e) Number of permutations of ‘n’ things, taken all at a time, when ‘m’ specified things always come together = n ! - [ m! x   (n-m+1)! ]

Example:   How many words can be formed with the letters of the word ‘OMEGA’ when:

(i)       ‘O’ and ‘A’ occupying end places.

(ii)       ‘E’ being always in the middle

(iii)       Vowels occupying odd-places

(iv)        Vowels being never together.

Ans.

(i)    When ‘O’ and ‘A’ occupying end-places

  => M.E.G. (OA)

  Here (OA) are fixed, hence M, E, G can be arranged in  3! ways

 But (O,A) can be arranged themselves is 2! ways.

=> Total number of words =  3!   x  2! = 12 ways.

(ii)  When ‘E’ is fixed in the middle

=>    O.M.(E), G.A.

Hence four-letter O.M.G.A. can be arranged in  4!   i.e 24 ways.

(iii)   Three vowels (O,E,A,) can be arranged in the odd-places (1^st, 3^rd and 5^th)    =  3!    ways.

And two consonants (M,G,) can be arranged in the even-place               (2^nd, 4^th) =   2 !   ways

=> Total number of ways= 3! x 2! = 12 ways.

(iv)  Total number of words   =   5!   =    120!

 If all the vowels come together, then we have: (O.E.A.), M,G

 These can be arranged in    3!    ways.

 But (O,E.A.) can be arranged themselves in   3! ways.

 => Number of ways, when vowels come-together  =    3!  x    3!  

= 36 ways

=> Number of ways, when vowels being never-together

= 120-36        =  84 ways.

Number of Combination of ‘n’ different things, taken ‘r’ at a time is given by:-

ⁿC_r=  n! / r ! x (n-r)!                                

Proof: Each combination consists of ‘r’ different things, which can be arranged among themselves in   r!  ways.

=> For one combination of ‘r’ different things, number of arrangements =    r!

For ⁿC_r combination number of arrangements:      r    ⁿC_r

=> Total number of permutations =    r!   ⁿC_r  ---------------(1) 

But number of permutation of ‘n’ different things, taken ‘r’ at a time

= ⁿP_r -------(2)

From (1) and (2) :

ⁿP_r  =      r!  .  ⁿC_r

or      n!/(n-r)!  =  r!   .  ⁿC_r          

or   ⁿC_r    =       n!/r!x(n-r)!

Note: ⁿC_r  =  ⁿC_n-r

or   ⁿC_r    = n!/r!x(n-r)!   and  ⁿC_n-r  =   n!/(n-r)!x(n-(n-r))!

 =  n!/(n-r)!xr!

Restricted – Combinations

(a)  Number of combinations of ‘n’ different things taken ‘r’ at a time, when ‘p’ particular things are always included = ^n-pC_r-p.

(b)  Number of combination of ‘n’ different things, taken ‘r’ at a time, when ‘p’ particular things are always to be excluded = ^n-pC_r

Example:      In how many ways can a cricket-eleven be chosen out of 15 players? if

(i)  A particular player is always chosen,

(ii)  A particular is never chosen.

Ans:   

(i)       A particular player is always chosen, it means that 10 players are selected out of the remaining 14 players.

=. Required number of ways =  ¹⁴C₁₀  = ¹⁴C₄

= 14!/4!x19!  = 1365                                              

(ii) A particular players is never chosen, it means that 11 players are selected out of 14 players.

 => Required number of ways =  ¹⁴C₁₁ 

 =   14!/11!x3!  = 364

(iii) Number of ways of selecting zero or more things from ‘n’ different things is given by:-   2ⁿ-1

Proof:  Number of ways of selecting one thing, out of n-things     = ⁿC₁

Number of selecting two things, out of n-things =ⁿC₂

Number of ways of selecting three things, out of n-things =ⁿC₃

Number of ways of selecting ‘n’ things out of ‘n’ things = ⁿC_n

=>Total number of ways of selecting one or more things out of n different things

= ⁿC₁ + ⁿC₂ + ⁿC₃ + ------------- + ⁿC_n

= (ⁿC₀ + ⁿC₁ + -----------------ⁿC_n)  - ⁿC₀

= 2ⁿ – 1                           [ ⁿC₀=1]

Example:  John has 8 friends. In how many ways can he invite one or more of them to dinner?

Ans.    John can select one or more than one of his 8 friends.

=> Required number of ways = 2⁸ – 1= 255.

(iv) Number of ways of selecting zero or more things from ‘n’ identical things is given by :- n+1

Example:   In how many ways, can zero or more letters be selected form the letters AAAAA?

Ans. Number of ways of :

         Selecting zero 'A's   =  1

         Selecting one 'A's   =   1

         Selecting two 'A's     = 1

         Selecting three 'A's  =  1

         Selecting four 'A's   =   1

         Selecting five 'A's    =  1

=> Required number of ways  = 6         [5+1]

(V)    Number of ways of selecting one or more things from ‘p’ identical things of one type ‘q’ identical things of another type, ‘r’ identical things of the third type and ‘n’ different things is given by :-

 (p+1) (q+1) (r+1)2ⁿ – 1

Example:    Find the number of different choices that can be made from 3 apples, 4 bananas and 5 mangoes, if at least one fruit is to be chosen.

Ans:

Number of ways of selecting apples = (3+1) = 4 ways.

Number of ways of selecting bananas = (4+1) = 5 ways.

Number of ways of selecting mangoes = (5+1) = 6 ways.

Total number of ways of selecting fruits = 4 x 5 x 6

But this includes, when no fruits i.e. zero fruits is selected

=> Number of ways of selecting at least one fruit = (4x5x6) -1  = 119

Note :- There was no fruit of a different type, hence here  n=o

 =>   2ⁿ  = 2⁰=1

(VI)   Number of ways of selecting ‘r’ things from ‘n’ identical things is ‘1’.

Example:   In how many ways 5 balls can be selected from ‘12’ identical red balls?

Ans. The balls are identical, total number of ways of selecting 5 balls  = 1.

Example: How many numbers of four digits can be formed with digits 1, 2, 3, 4 and 5?

Ans. Here n = 5                     [Number of digits]

And   r = 4                     [ Number of places to be filled-up]

Required number is   ⁵P₄ = 5!/1!  = 5 x 4 x 3 x 2 x 1

(a)  Number of combinations of ‘n’ different things taken ‘r’ at a time, when ‘p’ particular things are always included = ^n-pC_r-p.

(b)  Number of combination of ‘n’ different things, taken ‘r’ at a time, when ‘p’ particular things are always to be excluded = ^n-pC_r

Example:      In how many ways can a cricket-eleven be chosen out of 15 players? if

(i)  A particular player is always chosen,

(ii)  A particular is never chosen.

Ans:   

(i)       A particular player is always chosen, it means that 10 players are selected out of the remaining 14 players.

=. Required number of ways =  ¹⁴C₁₀  = ¹⁴C₄

= 14!/4!x19!  = 1365                                              

(ii) A particular players is never chosen, it means that 11 players are selected out of 14 players.

 => Required number of ways =  ¹⁴C₁₁ 

 =   14!/11!x3!  = 364

(iii) Number of ways of selecting zero or more things from ‘n’ different things is given by:-   2ⁿ-1

Proof:  Number of ways of selecting one thing, out of n-things     = ⁿC₁

Number of selecting two things, out of n-things =ⁿC₂

Number of ways of selecting three things, out of n-things =ⁿC₃

Number of ways of selecting ‘n’ things out of ‘n’ things = ⁿC_n

=>Total number of ways of selecting one or more things out of n different things

= ⁿC₁ + ⁿC₂ + ⁿC₃ + ------------- + ⁿC_n

= (ⁿC₀ + ⁿC₁ + -----------------ⁿC_n)  - ⁿC₀

= 2ⁿ – 1                           [ ⁿC₀=1]

Example:  John has 8 friends. In how many ways can he invite one or more of them to dinner?

Ans.    John can select one or more than one of his 8 friends.

=> Required number of ways = 2⁸ – 1= 255.

(iv) Number of ways of selecting zero or more things from ‘n’ identical things is given by :- n+1

Example:   In how many ways, can zero or more letters be selected form the letters AAAAA?

Ans. Number of ways of :

         Selecting zero 'A's   =  1

         Selecting one 'A's   =   1

         Selecting two 'A's     = 1

         Selecting three 'A's  =  1

         Selecting four 'A's   =   1

         Selecting five 'A's    =  1

=> Required number of ways  = 6         [5+1]

(V)    Number of ways of selecting one or more things from ‘p’ identical things of one type ‘q’ identical things of another type, ‘r’ identical things of the third type and ‘n’ different things is given by :-

 (p+1) (q+1) (r+1)2ⁿ – 1

Example:    Find the number of different choices that can be made from 3 apples, 4 bananas and 5 mangoes, if at least one fruit is to be chosen.

Ans:

Number of ways of selecting apples = (3+1) = 4 ways.

Number of ways of selecting bananas = (4+1) = 5 ways.

Number of ways of selecting mangoes = (5+1) = 6 ways.

Total number of ways of selecting fruits = 4 x 5 x 6

But this includes, when no fruits i.e. zero fruits is selected

=> Number of ways of selecting at least one fruit = (4x5x6) -1  = 119

Note :- There was no fruit of a different type, hence here  n=o

 =>   2ⁿ  = 2⁰=1

(VI)   Number of ways of selecting ‘r’ things from ‘n’ identical things is ‘1’.

Example:   In how many ways 5 balls can be selected from ‘12’ identical red balls?

Ans. The balls are identical, total number of ways of selecting 5 balls  = 1.

Example: How many numbers of four digits can be formed with digits 1, 2, 3, 4 and 5?

Ans. Here n = 5                     [Number of digits]

And   r = 4                     [ Number of places to be filled-up]

Required number is   ⁵P₄ = 5!/1!  = 5 x 4 x 3 x 2 x 1

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 WHAT HAS  HAPPENED TO OUR COUNTRY.............
THE COUNTRY WHICH HAS BEEN FAMOUS FOR HER CULTURE AND HERITAGE AND LOVE FOR OTHERS.
THAT COUNTRY IS NOW FACING THE MOST DANGEROUS TUMOUR IN THE WORLD....TERRORISM
SO LET US JOIN OUR HANDS & FIGHT AGAINST IT........!!!!!!!!!
JAI HIND

 Circular Motion



The planet Saturn and its well-known rings as viewed by telescope from Earth. The rings consists of particles, some very fine, some the size of boulders, which orbit the planet. Although their motion is not perfectly circular, all the important characteristics of the motion can be described by mathematics developed for describing circular motion.
With the subject of motion in more than one dimension broached, we can consider things other than simple linear motion. For example, many types of motion involve curved paths rather than linear ones. We have already seen one example in the parabolic shape of the trajectory of an object in free-fall. Another example is simply the motion of an object in a circular path. This includes motions as diverse as cars on a Ferris wheel to satellites in orbit around the earth. This kind of motion is somewhat special in that the object of interest maintains a constant speed as it goes around the circle, i.e. the magnitude of its velocity doesn't change. However, the direction of the velocity does have to change continuously to maintain the circular path. Whenever the velocity magnitude, direction, or both is changed, there must be a non-zero acceleration acting to change the motion. Look at the figure below. We've already discussed radians and their connection to radii and arc-lengths. Let's review how they would apply to finding the acceleration of an object in a circular path by looking at the figure below.

The object moves in a circular path of radius R. At some time, t_0, the velocity has direction shown in the figure and magnitude v. Some short time later, at t_1, the object has a velocity with a new direction but the same old magnitude, v. The angular position of the object has changed by an amount . The velocity change is, as shown, . The distance traveled by the object in time  is the arc-length . We can now start to relate the items we have defined. First, notice that, by the definition of radians, we have

where we have only talked about magnitudes. The direction for  we get by noting that it is approximately perpendicular to  by our picture. Since both  and  are equal to , then they are equal to each other, so

The definition of average acceleration during the time  is , so

where we remember that the magnitude of the objects velocity, v, and the radius R are constant in time. The time taken for the object to travel arc-length  is set by the time difference and the velocity, so we know that . Therefore, our average acceleration is just

As usual, we move from the average acceleration (which is just the average rate of change of the velocity) to the exact instantaneous acceleration by taking the limit as . This is the same as letting . In that case, the change in velocity is exactly perpendicular to . But since  is tangent to the circular path and therefore perpendicular to R, then the change in velocity is parallel or anti-parallel toR. From our picture, we see that  is anti-parallel (it points in toward the center while the radius points outward). Therefore, the acceleration needed to maintain a perfectly circular path of radius R with constant speed is

with direction inward toward the center of the circular path. When such a condition prevails, we say that the object is subject to a centripetal acceleration and that the object is undergoing uniform circular motion.
Of course, for an acceleration to be acting, there must be a force. Any force that provides an acceleration satisfying the centripetal condition (i.e. center-pointing, magnitude constant at ) is a centripetal force. A string tied to a rock whirled at constant velocity in a horizontal circle provides a centripetal force through the tension supplied by your hand. For the Ferris wheel, the tension is provided by struts which connect the seats to the wheel and the wheel to its center. Satellites are maintained in orbit by earth's gravity.