This problem is the generalization of exercise 2.3.9 from 'linear algebra' Hoffman and Kunze.

Let the set of vectors be linearly independent. Prove that are also linearly independent given

I mean

and so on.

ps: I proved it by proving a certain matrix to be invertible and that was pretty cumbersome. Hoping for a better pretty solution :)

## Tuesday, October 30, 2007

## Tuesday, October 23, 2007

### A closed form for nth Fibonacci number

I have made an attempt to arrive at a closed form by setting up a bijection in a counting process.

Consider the set . In how many ways can we select non null subset of such that no two elements in the subset give difference one ?

Let be the number of ways of selecting such a subset. Depending upon whether is included in the subset or not , we get

with and

It is clear that = (n+1)th Fibonacci number,

Looking the other way, let our subset contain elements.

{min, max}= {, }

I arrange all the elements in ascending order from left to right. Except for the last element, I place a ball just right of each element. Now I am left with balls with places available. You are getting what I am telling ?

This can be done in ways.

So,

=

Consider the set . In how many ways can we select non null subset of such that no two elements in the subset give difference one ?

Let be the number of ways of selecting such a subset. Depending upon whether is included in the subset or not , we get

with and

It is clear that = (n+1)th Fibonacci number,

Looking the other way, let our subset contain elements.

{min, max}= {, }

I arrange all the elements in ascending order from left to right. Except for the last element, I place a ball just right of each element. Now I am left with balls with places available. You are getting what I am telling ?

This can be done in ways.

So,

=

Labels:
enumerative combinatorics

### The Vodafone problem

I was lazily lying on my bed thinking soon after getting up still drowsy...I have change my hutch[no vodafone] account from postpaid to prepaid. Then this thing flashed to me:

1. Now vodafone announces a scheme by which you may choose number/s and you can make calls to that number/s at concessional rate. Lets call this condition as you two are 'connected'.You choosing that number/s also means that person [ those people] can also call you at concessional rate ie., automatically 'doubly connected'. Virtually,There is no limit to the number of people you can stay connected with. I choose some 'n' arbitrary distinct people and see who are connected to whom. I write (a->b) if a and b are connected. Here is a possible map:

1->2,4

2->1,5,6

3-> nothing

.

.

.

n->15,24

How many such maps are possible?

What if remove the condition that if (a->b) does not necessarily imply (b->a) ?

2. Consider this 'special scheme' in which if (a->b) and (b->c), then (a->c). [ I know this reminds you of equivalence relation :) ]. Now how many maps under both the conditions ?

1. Now vodafone announces a scheme by which you may choose number/s and you can make calls to that number/s at concessional rate. Lets call this condition as you two are 'connected'.You choosing that number/s also means that person [ those people] can also call you at concessional rate ie., automatically 'doubly connected'. Virtually,There is no limit to the number of people you can stay connected with. I choose some 'n' arbitrary distinct people and see who are connected to whom. I write (a->b) if a and b are connected. Here is a possible map:

1->2,4

2->1,5,6

3-> nothing

.

.

.

n->15,24

How many such maps are possible?

What if remove the condition that if (a->b) does not necessarily imply (b->a) ?

2. Consider this 'special scheme' in which if (a->b) and (b->c), then (a->c). [ I know this reminds you of equivalence relation :) ]. Now how many maps under both the conditions ?

Labels:
enumerative combinatorics

## Wednesday, October 10, 2007

### An intriguing one -II

This is the continuation of the post 'An intriguing one'. Here I make an attempt to look at a possible approach.

Consider two concentric circles with radii and . It is clear that there cannot exist two points within the b-circle.

I go on placing points on the circumference of a-circle with gap of at least 'b' of course, I do not know what to do with . As of now we shall distribute evenly. This was to give an idea. You must have figured out that we can pack still more, it is not difficult to see that we can place

points. So the picture now looks like fig1.

I have written only two circles with centers and respectively, to show available region for further points. After doing this, pick points like and in the next iteration. But analysis becomes non trivial from this step itself.

Of course the problem is solved of if points like lie within b-circle.

Miscellany: Seemingly related but a easy one i found at Colorado mathematical Olympiad

(a) We need to protect from the rain a cake that is in the shape of an equilateral triangle of side 2.1. All we have are identical tiles in the shape of an equilateral triangle of side 1. Find the smallest number of tiles needed.

(b) Suppose the cake is in the shape of an equilateral triangle of side 3.1. Will 11 tiles be enough to protect it from the rain?

I found that we require 6 and 11 .

Consider two concentric circles with radii and . It is clear that there cannot exist two points within the b-circle.

I go on placing points on the circumference of a-circle with gap of at least 'b' of course, I do not know what to do with . As of now we shall distribute evenly. This was to give an idea. You must have figured out that we can pack still more, it is not difficult to see that we can place

points. So the picture now looks like fig1.

I have written only two circles with centers and respectively, to show available region for further points. After doing this, pick points like and in the next iteration. But analysis becomes non trivial from this step itself.

Of course the problem is solved of if points like lie within b-circle.

Miscellany: Seemingly related but a easy one i found at Colorado mathematical Olympiad

(a) We need to protect from the rain a cake that is in the shape of an equilateral triangle of side 2.1. All we have are identical tiles in the shape of an equilateral triangle of side 1. Find the smallest number of tiles needed.

(b) Suppose the cake is in the shape of an equilateral triangle of side 3.1. Will 11 tiles be enough to protect it from the rain?

I found that we require 6 and 11 .

Labels:
general

## Sunday, October 07, 2007

### Pleasure of Latex

Many many thanks to Wolverine for latex implementation script on Firefox.

Link to the site.

good naa.. :)

ps: But I found that same cant be used on be comments page. Anybody has a idea?

Link to the site.

good naa.. :)

ps: But I found that same cant be used on be comments page. Anybody has a idea?

Labels:
household

### Distribution with restrictions [S]

When I wrote this post I made a typo mistake. Here is what I intended.

*******************************************************************

Problem: Enumerate the number of solutions of the equation

such that for every

Lets suppose so that at least one solution is assured.

*************************

The flipside of this is simple. How many solutions with each

? Easy.. put balls in each of the boxes, a priori. Now distribute rest in :)

***********************************************************************

solution: Suppose

Hence the number of solutions with the given restriction is

*******************************************************************

Problem: Enumerate the number of solutions of the equation

such that for every

Lets suppose so that at least one solution is assured.

*************************

The flipside of this is simple. How many solutions with each

? Easy.. put balls in each of the boxes, a priori. Now distribute rest in :)

***********************************************************************

solution: Suppose

Let

(i) only one of the entries has size

and rest are less than that. Now we get the regular inclusion-exclusion expression(ii) For two,

and so on.

Summing up individually we get

Hence the number of solutions with the given restriction is

Labels:
enumerative combinatorics

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