Factorial

Factorial

The factorial of a positive integer n, denoted by n!, is the product of all positive integers less than or equal to n.

For example 5! = 1*2*3*4*5 = 120

Remember it  0! = 1
                       n! = n*(n-1)!
Basically the question came to competition.
Qus.1..What is the maximum power of 2 in 50!
             Or
   How many 2 in 50!
Solution

50! = 1*2*3*4*5*6*7*8*9*10……………………………………………….50

50! = 1*2*3*4*5*6*7*8*9*10……………………………………………….50

Since every alternate number is multiple of 2

That is twenty five(25) times of 2

Again in

50! = 1*2*3*4*5*6*7*8*9*10……………………………………………….50

Every multiple of 4 is at least two times 2

That is twelve(12) times of 2

Again in

50! = 1*2*3*4*5*6*7*8*9*10……………………………………………….50

Every multiple of 8 is at least two times 3

That is six(6) times of 2

Again in

50! = 1*2*3*4*5*6*7*8*9*10……………………………………………….50

Every multiple of 16 is at least two times 4

That is three(3) times of 2

Again in

50! = 1*2*3*4*5*6*7*8*9*10……………………………………………….50

Every multiple of 32 is at least two times 5

That is one(1) times of 2

Hence 25+12+6+3+1 = 47

Hence there are 47 time 2 in 50!

Qus.2.What is the maximum power of 5 in 50!
             Or
   How many 5 in 50!
Solution

Qus.1..What is the maximum power of 6 in 50!
             Or
   How many 6 in 50!
Solution        6=2*3
                there are minimum of power 2 and 3

Hence there are 22 time 6 in 50!

Logical Venn Diagrams

Logical Venn Diagrams
Venn Diagrams :-

                    A Venn diagram  or logic diagram is a diagram or Euler Venn diagram  
                    that shows all possible

                  logical relations between a finite collection of different sets.Logical

                   A Venn diagram is the area of each shape is proportional to the number of

                   elements it contains is called an area-proportional or scaled Venn diagram.

In Other words

                    It is a process of shows relationship between  various geometric strictures.

                    Intersection between two geometric structures indicate that they have

                    something in common .
Representation of Venn diagram
                             (1)

                           (2)

Basic Formula of Venn diagram
                     (1). n(A⋃B) = n(A) +n(B) -n(A⋂B)
                     (2).n(A⋃B⋃C) = n(A) +n(B)+n(C) -n(A⋂B)- n(B⋂C)- n(B⋂C)+n(A⋂B⋂C)

Examples 1.The triangle circle and rectangle shown below representation smart ,

                      hardworking and educated persons  respectively  which one the areas marked

                      1 to 6 is represented by the hardworking educated persons who are not smart. 

                    

                It is clear that the required region is the one which is common in circle and rectangle

                but not in triangle i.e. region 5 which shows in co lour  red .

Examples 2.The rectangle triangle circle rectangle and square shown below representation

                     males ,educated,urban and civil services.

                    Answers the following questions

                   (1). How many peoples are educated male not an urban resident.

                   (2).How many are educated male an urban resident. 

Ans. (1). 11            (2).4

Examples 3.

                   The  triangle circle  and square shown below representation

                     educated people,employed people backward people .

                    Answers the following questions

                   (1). How many backward people are educated.

                   (2).How many backward people are not educated.. 

            

Ans. (1). 14            (2).22.

Examples 4.

                  The circle represents players

                  The triangle represents outdoor games

                  The rectangle represents indoor games

                  The square represents national level  players

Answer the following

(1) Which letter represent the players who play indoor games at national level.

(2) Which letter represent the outdoor game as well as indoor game players who

       do not  play at national level.

Examples 5.In a class of 35 students, 24 like to play cricket and 16 like to play football. Also,

                    each student likes to play at least one of the two games. How many students like to

                    play both cricket and football

Explanation :-Let C denote student play cricket,F denote student play football,

                       Then n(C)=24,n(F)=16,n(C⋃F)= 35  ,n(C⋂F) = ?

                       By the formula

                      n(A⋃B) = n(A) +n(B) -n(A⋂B)

                       n(C⋃F) = n(C) + n(F) - n(C⋂F)

                        35 = 24 + 16 -n(C⋂F)

                      n(C⋂F) = 40 - 35 = 5

                      Hence  5  students like to play both cricket and football.

Examples 6 A sport club  awarded 38 medals in handball, 15 in basketball and 20 in football. 

                   If these medals went to a total of 58 men and only three men got medals in

                   all the three sports, how many received medals in exactly two of the three sports ?

Explanation :-Let H denote  men play handball ,B denote men  play  basketball ,

                         F denote men  play  football,

                       Then n(H)=38,n(B)=15, n(F)=20,

                      n(H⋃B⋃F) = 58 ,n(H⋂B⋂F) = 3

                         n (H∪ B ∪ F ) = n ( H ) + n ( B ) + n ( F ) – n (H ∩ B ) – n (B ∩ F ) – n (F ∩ H ) + n (H∩ B ∩ F ), 

                          n (H ∩ B ) + n (B ∩ F ) +n (F ∩ H ) = n ( H ) + n ( B ) + n ( F ) + n (H∩ B ∩ F )-n (H∪ B ∪ F )

.                                                                      = 38 +15 + 20 +3 -58

                                                                       =  18 (Red Shaded region)

Counting of Figures

Counting of Figures
(1) Counting of Triangles:-

In the counting of triangles, we have a figure. From the given figure we will have to

 identify how many triangles in a given known figure . 

 TYPE 1:-
Example 1. consider the following figure

                     There are 3 triangles in figure.

                     In single horizontal line sum of inside triangles 1+2=3

Example 2. consider the following figure

                    There are 6 triangles in figure.1,2,3,4,1+2,2+3

                    In single horizontal line sum of inside triangles 1+2+3=6

Example 3. consider the following figure

      There are 10 triangles in figure.1,2,3,4,1+2,2+3,3+4,1+2+3,2+3+4,(1+2+3+4) greater triangle

                    In single horizontal line sum of inside triangles 1+2+3+4=10

 TYPE 2:-

Example 4. consider the following figure

There are 6 triangles in figure.1,2,1+2,1+3,2+4,(1+2+3+4) greater triangle

                    In two horizontal lines sum of inside triangles 2(1+2)=6.

Example 5. consider the following figure

There are 20 triangles in figure.2✖{1,2,3,4,1+2,2+3,3+4,1+2+3,2+3+4,(1+2+3+4) greater triangle} 

                    In two horizontal lines sum of inside triangles 2(1+2+3+4)=20.

Example 6. consider the following figure

There are 30 triangles in figure.3✖{1,2,3,4,1+2,2+3,3+4,1+2+3,2+3+4,(1+2+3+4) greater triangle} 

                    In three horizontal lines sum of inside triangles 3(1+2+3+4)=30.

 TYPE 3:-

Example 7. consider the following figure

There are 5 triangles in figure.

Triangle inside triangle 1 hence number of triangles = 4✖1 +1 (greater triangle) = 5 

Example 8. consider the following figure

There are 9 triangles in figure.

Triangle inside triangles 2 hence number of triangles = 4✖2 +1 (greater triangle) = 9 

Example 9. consider the following figure

There are 13 triangles in figure.

Triangle inside triangles 3 hence number of triangles = 4✖3 +1 (greater triangle) = 9 

TYPE 4:-

Example 10. consider the following figure

There are 11 triangles in figure.

Triangle inside triangle 1 hence number of triangles = 4✖1 +1 (greater triangle) = 5 

and triangles 6 are (1,2,3,4,5,6) 

Hence total triangles 5+6 = 11

Example 11. consider the following figure

There are 15 triangles in figure.

Triangle inside triangle 2 hence number of triangles = 4✖2 +1 (greater triangle) = 9 

and triangles 6 are (1,2,3,4,5,6) 

Hence total triangles 9+6 = 15

TYPE 5:-

Example 12. consider the following figure

Triangles 1,2,3,4,1+2,2+3,3+4,4+1

i.e. 4✖2=8

There are 8 triangles

Example 13. consider the following figure

Example 14. consider the following figure

Example 15. consider the following figure

Example 16. consider the following figure

Example 17. consider the following figure

Example 18. consider the following figure

Example 19. consider the following figure

Example 20. consider the following figure

(2) Counting of Square:-

In the counting of squares, we have a figure. From the given figure we will have to

 identify how many squares in a given known figure . 
Type-1
If number of rows = number of columns =n
Then number of squares = ∑n²

Example 1.:-consider the following figure

Type-2
If number of rows = n,number of columns =m

Then number of squares = m✖n +(m-1)✖(n-1) + ............

Example 2.:-consider the following figure

Type-3

Example 3. :-consider the following figure

Alligation and Mixture

 Alligation and Mixture

                            Alligation is the rule that find the ratio in which two or more ingredients 

                            at the given price must be mixed to produce a mixture of a desired price.

Mean Price :-The cost price of unit quantity of the mixture.

Rule of alligation :-If two ingredients are mixed,then

      Cheaper quantity : Dearer quantity = (D-M) : (M-C)                

Replacement of part of mixture :-Suppose a container contains a solution from which
                                                     some quantity of solution is taken out and replaced with
                                                    one of the ingredients.This process is repeated n times then  

Example 1.:-In what ratio must rice at Rs.40 /kg to mix rice at Rs.60 /kg,so that  the average cost of mixture is Rs.55 /kg ?

Explanation :-

Hence the answer is 60-55 : 55-40 = 5:15

                                                             1:3

 Example 2.:-

In what ratio must a grocer mix two varieties of pulses costing Rs. 15 and  

                       Rs. 20 per kg respectively so as to get a mixture worth Rs. 16.50 kg?

Explanation :-

              

Hence the answer is 20-16.50:16.50-15=3.50 :1.50

                                                             7:3

Example 3.:-How many kilogram of sugar costing Rs. 9 per kg must be mixed with 27 kg

                      of sugar costing Rs. 7 per kg so that there may be a gain of 10% by selling

                      the mixture at Rs. 9.24 per kg?

Explanation :-S.P. of 1 kg of mixture = Rs. 9.24, 

                        Gain 10%.

C.P. of 1 kg of mixture = Rs.		100	x 9.24		= Rs. 8.40
		110

By the rule of alligation, we have:

           Ratio of quantities of 1^st and 2^nd kind = 1.40 : 0.60 = 7 : 3.

           Let x kg of sugar of 1^st be mixed with 27 kg of 2^nd kind.

          Then, 7 : 3 = x : 27

               x = (27✖7)/3 = 63

Example 4.:-Two containers A and B contain milk and water in the ratio of 5 : 2 and 

                     7 : 6 respectively. Find the ratio in which these two mixtures can be mixed

                     so that a new mixture formed in the container C is in the ratio of 8 : 5. 

Explanation :-Let the cost price of milk be Rs. 1 per liter.

              Therefore, cost of milk in 1 liter of mixture in

Container A (Milk : Water = 5 : 2) =5/7 ✖Rs.1

=5/7

Container B (Milk : Water = 7 : 6) =

7/13✖Rs.1 = 7/13

Container C (Milk : Water = 8 : 5) =

8/13✖Rs.1 = 8/13

             Now by alligation  rule , 

The required ratio of milk and water = 1/13 : 9/91
                                                               7 : 9

Example 5.:-A container contains 40 litres of milk. From this container 4 litres of milk

                      was taken out and replaced by water. This process was repeated further

                      two times. How much milk is now contained by the container?

Explanation :-

Example 6.:-8 liters are drawn from a  container full of milk. then filled with water

                        This process was repeated three more times. The ratio of the quantity of

                        milk now left in container to that of water is 16:65.How much milk did

                        the container hold originally?

Explanation :-Let the quantity of milk in the container is x liters.

                         Then  the quantity of milk in the container after 4 operations 
                                                                           

                                                                    x = 24 liters