A note on Permutations and Combinations

Here is a situation where one answer has many questions.






 





...Try to add to the list!!!...are these independent of each other?

This should challenge you!!!

Delete 100 digits in the number

12345678910111213141516171819. . . . 96979899100

so that the resultant number has

(a)  The greatest possible value;
(b)  The least possible value.

Solve this!!!

For x > 1 determine the sum of the infinite series

One more for you to try

For

Show that



Now, for

and

Show that



Finally, show that the limit

Try this!!!

In how many ways can a cube be painted with six different colours with one colour on each face?

Use of complex numbers in Integration

Twin Integrals :
To find

and


We begin with

Let k = a + ib
then,



Equating Real and Imaginary parts

and

Conic Sections and Eccentricity(e)

Let us examine the process of Limit e ---> 1of a conic

 The X shaped lines that define the surface of the cone are called generators and the vertical line passing through the intersection of the generator is called axis 
 
We have conic sections formed when a double right circular cone is intersected by a plane.  
General equation of a conic is  b²x²+a²y² = a²b²

where, b² = a²|1-e²|

When e = 0 , we get the conic section- 'Circle'
Note: Here the plane cuts the double cone at right angle with the axis of the cone

  When e < 1 , we get the conic section- 'Ellipse'

Note: Here the plane cuts the double cone at an inclination with the axis of the cone

 When e = 1 , we get the conic section- 'Parabola' 
Note: Here the plane lies parallel to the generator of the double cone

 When e > 1 , we get the conic section- 'Hyperbola'
Note : Here the plane cuts bottom and top cones of the double cone

 So as Eccentricity(e) varies the conic section varies as shown below

Key Note: Can we conclude the process in the following form

when e = 0 we have a perfectly symmetrical figure which is a circle. 

As e moves from 0 to 1, the circle gets deformed. As e approaches 1 the deformation becomes deeper and deeper and as e equals 1 there is an explosion instantaneously and the conic section becomes a parabola. Further as e moves from 1 instaneously another explosion results into a hyperbola.

Find the value of 1² - 2² + 3² - 4² + 5²……… - 2010² + 2011²

The required solution can be obtained using the following approaches                         

        Application of the identity a² - b² = (a + b)(a - b)

1² - 2² + 3² - 4² + 5²……… - 2010² + 2011²

= 2011² - 2010²…. +5² - 4² + 3² - 2² + 1²

= (2011 - 2010)(2011 + 2010) +……..+ (3 - 2)(3 + 2) + 1

= (1).(2011 + 2010)+…….+ (1).(3 + 2)+1

= 2011 + 2010 + …….. +3 +2 +1

 
 Method of generalisation

1² - 2² + 3² = 1 + 2 + 3

1² - 2² + 3² - 4² + 5²  = 1 + 2 + 3 + 4 + 5

Hence it follows that 1² - 2² + 3² - 4² + 5².... - 2010² + 2011²

is equal to 1 + 2 + 3 + 4 + 5 +…..+ 2010 + 2011

 

Reusability or Generalisation of a result in problem solving-3

Let us see another illustration of generalisation of a result

Problem Description : Two circles of radii R and r touch each other externally. Find the radius r1 of circle touching both these externally and their direct common tangent. Also find the radius r2 of circle which touches the second and third circles and their direct common tangent.


Solution : 

From the above figure, we get the radius r1 as

Similarly, we get r2 as

Now by substituting 

we get

Generalisation : Proceeding like this the radius rn of nth such circle is

Simple application of |x|

Square root of 4 is

Positive Square root of 4 is
2

Positive square root of x is denoted by


In general,


When we say IN WORDS "Square root of 4" it means


However when we say "Positive square root of 4" or symbolically it equals 2

Reusability or Generalisation of a result in problem solving-2

Let us now see a problem that focus on generalisation of a result to solve a problem

Problem Description : Box A1 contains a white balls and b black balls. Box A2 has a white balls and b black balls. A ball is drawn from A1 and is transferred to A2. Now, if a ball is drawn from A2 find the probability that it is white.


Solution  :  Upon observation, we can see that the required answer can be easily obtained by the application of Bayes' Theorem as

Generalisation :
    A1, A2, A3 .... An are n boxes each containing a white balls and b black balls. A ball is transferred from A1 to A2 and then from A2 to A3 and so on such that the last transfer is from An-1 to An. Then the probability that a ball drawn from An is white is 

 

Reusability or Generalisation of a result in problem solving

Problem Description :  Find area of the shaded region ABCD in the figure below.

Solution :

 Observation : We begin with the area of the shaded region as shown in the above figure which can be obtained easily using integration as

Now the required Ar(ABCD) = AR(OBCO) - AR(DOAD) - AR(BAOB) + AR(AOA) where each of the areas of regions OBCO, DOAD, BAOB, AOA can be obtained by reusing the above observation accordingly.


Let us now see a similar illustration of the reusability of a result


Problem Description : Find the area of ∆leABC in the below figure

Solution :


Observation : Consider the ∆le PQR as shown in the figure below

We have,

Therefore,

Now, from the observation,
the required Ar(∆leABC) =AR(∆leAFE) - AR(∆leBFD) +AR(∆leCED)