Prof. Tim Gowers’ on recognising countable sets

https://gowers.wordpress.com/2008/07/30/recognising-countable-sets/

Thanks Dr. Gowers’. These are invaluable insights into basics. Thanks for giving so much of your time.

A Primer: Generating Functions: Part I : RMO/INMO 2019

GENERATING FUNCTIONS and RECURRENCE RELATIONS:

The concept of a generating function is one of the most useful and basic concepts in the theory of combinatorial enumeration. If we want to count a collection of objects that depend in some way on n objects and if the desired value is say, $\phi (n)$ , then a series in powers of t such as $\sum \phi (n) t^{n}$ is called a generating function for $\phi (n)$ . The generating functions arise in two different ways. One is from the investigation of recurrence relations and another is more straightforward: the generating functions arise as counting devices, different terms being specifically included to account for specific situations which we wished to count or ignore. This is a very fundamental, though difficult, technique in combinatorics. It requires considerable ingenuity for its success. We will have a look at the bare basics of such stuff.

We start here with the common knowledge:

$(1+\alpha_{1}t)(1+\alpha_{2}t)\ldots (1+\alpha_{n}t)=1+a_{1}t+a_{2}t^{2}+ \ldots + a_{n}t^{n}$ ….(2i) where $a_{r}=$ sum of the products of the $\alpha$ ‘s taken r at a time. …(2ii)

Incidentally, the $a$ ‘s thus defined in (2ii) are called the elementary symmetric functions associated with the $a$ ‘s. We will re-visit these functions later.

Let us consider the algebraic identity (2i) from a combinatorial viewpoint. The explicit expansion in powers of t of the RHS of (2i) is symbolically a listing of the various combinations of the $\alpha$ ‘s in the following sense:

$a_{1}=\sum \alpha_{1}$ represents all the 1-combinations of the $\alpha$ ‘s
$a_{2}=\sum \alpha_{1}\alpha_{2}$ represents all the 2-combinations of the $\alpha$ ‘s
and so on.

In other words, if we want the r-combinations of the $\alpha$ ‘s, we have to look only at the coefficients of $t^{r}$ . Since the LHS of (2i) is an expression which is easily constructed and its expansion generates the combinations in the said manner,we say that the LHS of (2i) is a Generating Function (GF) for the combinations of the $\alpha$ ‘s. It may happen that we are interested only in the number of combinations and not in a listing or inventory of them. Then, we need to look for only the number of terms in each coefficient above and this number will be easily obtained if we set each $\alpha$ as 1. Thus, the GF for the number of combinations is $(1+t)(1+t)(1+t)\ldots (1+t)$ n times;

and this is nothing but $(1+t)^{n}$ . We already know that the expansion of this gives $n \choose r$ as the coefficient of $t^{r}$ and this tallies with the fact that the number of r-combinations of the $\alpha$ ‘s is $n \choose r$ . Abstracting these ideas, we make the following definition:

Definition I:
The Ordinary Generating Function (OGF) for a sequence of symbolic expressions $\phi(n)$ is the series

$f(t)=\sum_{n}\phi (n)t^{n}$ …(2iii)

If $\phi (n)$ is a number which counts a certain type of combinations or permutations, the series $f(t)$ is called the Ordinary Enumeration (OE) or counting series for $\phi (n)$ for $n=1,2,\ldots$

Example 2:
The OGF for the combinations of five symbols a, b, c, d, e is $(1+at)(1+bt)(1+ct)(1+dt)(1+et)$

The OE for the same is $(1+t)^{5}$ . The coefficient of $t^{4}$ in the first expression is

(*) abcd+abce+ abde+acde+bcde.

The coefficient of $t^{4}$ in the second expression is $5 \choose 4$ , that is, 5 and this is the number of terms in (*).

Example 3:

The OGF for the elementary symmetric functions $a_{1}, a_{2}, \ldots$ in the symbols $\alpha_{1},\alpha_{2}, \alpha_{3}, \ldots$ is $(1+\alpha_{1}t)(1+\alpha_{2}t)(1+\alpha_{3}t)\ldots$ ….(2iv)

This is exactly the algebraic result with which we started this section.

Remark:

The fact that the series on the HRS of (2iii) is an infinite series should not bother us with questions of convergence and the like. For, throughout (combinatorics) we shall be working only in the framework of “formal power series” which we now elaborate.

*THE ALGEBRA OF FORMAL POWER SERIES*

The vector space of infinite sequences of real numbers is well-known. If $(\alpha_{k})$ and $\beta_{k}$ are two sequences, their sum is the sequence $(\alpha_{k}+\beta_{k})$ , and a scalar multiple of the sequence $(\alpha_{k})$ is $(c\alpha_{k})$ . We now identify the sequence $(\alpha_{k})$ with $k=0,1,2, \ldots$ with the “formal” series

$f = \sum_{k=0}^{\infty}\alpha_{k}t^{k}$ ….(2v)

where $t^{k}$ only means the following:

$t^{0}=1$ , $t^{k}t^{l}=t^{k+l}$ .

In the same way, $(\beta_{k})$ , where $k=0,1,2,\ldots$ corresponds to the formal series:

$g=\sum_{k=0}^{\infty}\beta_{k}t^{k}$ and

we define: $f+g = \sum (\alpha_{k}+\beta_{k})t^{k}$ , and $cf= \sum (c\alpha_{k})t^{k}$ .

The set of all power series f now becomes a vector space isomorphic to the space of infinite sequences of real numbers. The zero element of this space is the series with every coefficient zero.

Now, let us define a product of two formal power series. Given f and g as above, we write $fg=\sum_{k=0}^{\infty}\gamma_{k} t^{k}$ where

$\gamma_{k}=\alpha_{0}\beta_{k}+\alpha_{1}\beta_{k-1}+\ldots + \alpha_{k}\beta_{0} = \sum (\alpha_{i}\beta_{j})$ , where $i+j=k$ .

The multiplication is associative, commutative, and also distributive wrt addition. (the students/readers can take up this as an appetizer exercise !!) In fact, the set of all formal power series becomes an algebra. It is called the algebra of formal power series over the real s. It is denoted by $\bf\Re[t]$ , where $\bf\Re$ means the algebra of reals. We further postulate that $f=g$ in $\bf\Re[t]$ iff $\alpha_{k}=\beta_{k}$ for all $k=0,1,2,\ldots$ . As we do in polynomials, we shall agree that the terms not present indicate that the coefficients are understood to be zero. The elements of $\bf\Re$ may be considered as elements of $\bf\Re[t]$ . In particular, the unity 1 of $\bf\Re$ is also the unity of $\bf\Re[t]$ . Also, the element $t^{n}$ with $n>0$ belongs to $\bf\Re$ , it being the formal power series $\sum \alpha_{k}t^{k}$ with $\alpha_{n}=1$ and all other $\alpha$ ‘s zero. We now have the following important proposition which is the only tool necessary for working with formal power series as far as combinatorics is concerned:

Proposition : 2_4:
The element f of $\bf\Re[t]$ given by (2v) has an inverse in $\bf\Re[t]$ iff $\alpha_{0}$ has an inverse in $\bf\Re$ .

Proof:
If $g=\sum \beta_{k}t^{k}$ is such that $fg=1$ , the multiplication rule in $\bf\Re[t]$ tells us that $\alpha_{0}\beta_{0}=1$ so that $\beta_{0}$ is the inverse of $\alpha_{0}$ . Hence, the “only if” part is proved.

To prove the “if” part, let $\alpha_{0}$ have an inverse $\alpha_{0}^{-1}$ in $\bf\Re$ . We will show that it is possible to find $g=\sum \beta_{k}t^{k}$ in $\bf\Re[t]$ such that $fg=1$ . If such a g were to exist, then the following equations should hold in order that $fg=1$ , that is,

$\alpha_{0}\beta_{0}=1$
$\alpha_{0}\beta_{1}+\alpha_{1}\beta_{0}=0$
$\alpha_{0}\beta_{2}+\alpha_{1}\beta_{1}+\alpha_{2}\beta_{0}=0$
$\vdots$

So we have $\beta_{0}=\alpha_{0}^{-1}$ from the first equation. Substituting this value of $\beta_{0}$ in the second equation, we get $\beta_{1}$ in terms of the $\alpha$ ‘s. And, so on, by the principle of mathematical induction, all the $\beta$ ‘s are uniquely determined. Thus, f is invertible in $\bf\Re$ . QED.

Note that it is the above proposition which justifies the notation in $\bf\Re[t]$ , equalities such as

$\frac{1}{1-t}=1+t+t^{2}+t^{3}+\ldots$

The above is true because the RHS has an inverse and $(1-t)(1+t+t^{2}+t^{3}+\ldots)=1$

So, the unique inverse of $1+t+t^{2}+t^{3}+\ldots$ is $(1-t)$ and vice versa. Hence, the expansion of $\frac{1}{1-t}$ as above. Similarly, we have

$\frac{1}{1+t}=1-t+t^{2}-\ldots$
$\frac{1}{1-t^{2}}=1+t^{2}+t^{4}+\ldots$ and many other such familiar expansions.

There is a differential operator in $D$ in $\bf\Re[t]$ , which behaves exactly like the differential operator of calculus.

Define: $(Df)(\alpha)=\sum_{k=0}^{\infty}(k+1)\alpha_{k+1}t^{k}$

Then, one can easily prove that $D: f \rightarrow Df$ is linear on $\bf\Re[t]$ , and further
$D^{r}f(t)=\sum_{k=0}^{\infty}(k+r)(k+r-1)\ldots(k+1)\alpha_{k+r}t^{k}$ from which we get the term “Taylor-MacLaurin” expansion

$f(t)=f(0)+Df(0)+\frac{D^{2}f(0)}{2!}+ \ldots$ …(2vi)

In the same manner, one can obtain, from $f(t)=\frac{1}{1-\alpha t}$ , which in turn is equal to
$1+ \alpha t + \alpha^{2} t^{2}+ \alpha^{3} t^{3} + \ldots$

the result which mimics the logarithmic differentiation of calculus, viz.,

$\frac{(Df)(t)}{f(t)} = \alpha + \alpha^{2} t+ \alpha^{3}t^{2}+ \alpha^{4}t^{3}+\ldots$ …(2vii)

The truth of this in $\bf\Re[t]$ is seen by multiplying the series on the RHS of (2vii) by the series for $f(t)$ , and thus obtaining the series for $(Df)(t)$ .

Let us re-consider generating functions now. We saw that the GF for combinations of $\alpha_{1}, \alpha_{2}, \ldots, \alpha_{n}$ is $(1+\alpha_{1}t)(1+\alpha_{2}t)\ldots(1+\alpha_{n}t)$ .

Let us analyze this and find out why it works. After all, what is a combination of the symbols : $\alpha_{1}, \alpha_{2}, \ldots, \alpha_{n}$ ? It is the result of a decision process involving a sequence of independent decisions as we move down the list of the $\alpha$ ‘s. The decisions are to be made on the following questions: Do we choose $\alpha_{1}$ or not? Do we choose $\alpha_{2}$ or not? $\ldots$ Do we choose $\alpha_{n}$ or not? And, if it is an r-combination that we want, we say “yes” to r of the questions above and say “no” to the remaining. The factor $(1+\alpha_{1}t)$ in the expression (2ii) is an algebraic indication of the combinatorial fact that there are only two mutually exclusive alternatives available for us as far as the symbol $\alpha_{1}$ is concerned. Either we choose $\alpha_{1}$ or not. Choosing “ $\alpha_{1}$ ” corresponds to picking the term $\alpha_{1}t$ and choosing “not $-\alpha_{1}$ ” corresponds to picking the term 1. This correspondence is justified by the fact that in the formation of products in the expression of (2iv), each term in the expansion has only one contribution from $1+\alpha_{1}t$ and that is either $1$ or $\alpha_{t}$ .

The product $(1+\alpha_{1}t)(1+\alpha_{2}t)$ gives us terms corresponding to all possible choices of combinations of the symbols $\alpha_{1}$ and $\alpha_{2}$ — these are:

$1.1$ standing for the choice “not- $\alpha_{1}$ ” and “not- $\alpha_{2}$ ”

$\alpha_{1}t . 1$ standing for the choice of $\alpha_{1}$ and “not- $\alpha_{2}$ ”

$1.\alpha_{2}t$ standing for the choice of “not- $\alpha_{1}$ ” and $\alpha_{2}$ .

$\alpha_{1}t . \alpha_{2}t$ standing for the choice of $\alpha_{1}$ and $\alpha_{2}$ .

This is, in some sense, the rationale for (2iv) being the OGF for the several r-combinations of $\alpha_{1}, \alpha_{2}, \ldots, \alpha_{n}$ .

We shall now complicate the situation a little bit. Let us ask for the combinations of the symbols $\alpha_{1}, \alpha_{2}, \ldots, \alpha_{n}$ with repetitions of each symbol allowed once more in the combinations.

To be discussed in the following article,

Regards,
Nalin Pithwa.

Reference:
Combinatorics, Theory and Applications, V. Krishnamurthy, East-West Press.
Amazon India Link:
https://www.amazon.in/Combinatorics-Theory-Applications-Krishnamurthy-V/dp/8185336024/ref=sr_1_5?keywords=V+Krishnamurthy&qid=1553718848&s=books&sr=1-5

Tutorial problems for RMO 2019 : combinatorics continued

Posted by Nalin Pithwa

1) In how many ways can 5 men and 5 women be seated in a round table if no two women may be seated side by side?

2) Six generals propose locking a safe containing top secret with a number of different locks. Each general will be given keys to certain of these locks. How many locks are required and how many keys must each general have so that, unless at least four generals are present, the safe cannot be opened?

3) How many integers between 1000 and 9999 inclusive have distinct digits? Of these, how many are even numbers? How many consist entirely of odd digits?

4) In how many ways can 9 distinct objects be placed in 5 distinct boxes in such a way that 3 of these boxes would be occupied and 2 would be empty?

5) In how many permutations of the word AUROBIND do the vowels appear in the alphabetical order?

6) There is an unlimited supply of weights of integral numbers of grams. Using n or fewer weights, find the number of ways in which a weight of m grams can be obtained. Prove that there is a bijection of the set of all such ways on the set of increasing words of length $(n-1)$ or $(m+1)$ ordered letters.

7) How many distinct solutions are there of $x+y+z+w=10$ (a) in positive integers and (b) in non-negative integers?

8) A train with n passengers aboard makes m stops. In how many ways can the passengers distribute themselves among these m stops as alighting passengers? if we are concerned only with the number of alighting passengers at each stop, how would the answer be modified?

9) There are 16 books on a bookshelf. In how many ways can 6 of these books be selected if a selection must not include two neighbouring books?

10) Show that there are ${(n=5)} \choose 5$ distinct throws of a throw with n non-distinct dice.

11) Given n indistinguishable objects and n additional distinct objects —- also distinct from the earlier n objects — in how many ways can we choose n out of the 2n objects?

12) Establish the following relations:
12a) $B_{n+1}=\sum_{k=0}^{n}(B_{k}){n \choose k}$
12b) $\sum_{k}{p \choose k}{q \choose {n-k}}={{p+q} \choose n}$
12c) $S_{n+1}^{m} = \sum_{k=0}^{n}{n \choose k}S_{k}^{m-1}$
12d) $n^{p}=\sum_{k=0}^{n}{n \choose k}k! (S_{p}^{k})$

13) Prove the following identity for all real numbers x:
$x^{n}= \sum_{k=1}^{n}S_{n}^{k}[x]_{k}$

14) Express $x^{4}$ in terms of ${x \choose 4}$ , ${x \choose 3}$ , …by using the $S_{n}^{k}$ ‘s. Express ${x \choose 4}$ in terms of $x^{4}$ , $x^{3}$ , …by using the $s_{n}^{k}$ ‘s.

15) A circular loop is divided into p parts, p prime. In how many ways can we paint the loop with n colours if we do not distinguish between patterns which differ only by a rotation of the loop? Deduce Fermat’s Little theorem: $n^{p}-n$ is divisible by p if p is prime.

16) In problem 15, prove that $n^{p}-n$ is also divisible by 2p if $p \neq 2$ . Where is the hypothesis that p is prime used in Problem 15 or in this problem?

17) How many equivalence relations are possible on an n-set?

18) The complete homogeneous symmetric function of n variables $\alpha_{1}$ , $\alpha_{2}$ , $\ldots$ , $\alpha_{n}$ of degree r is defined as $h_{r}(\alpha_{1},\alpha_{2}, \alpha_{3}, \ldots, \alpha_{n})=\sum \alpha_{1}^{i_{1}}\alpha_{2}^{i_{2}}\ldots \alpha_{n}^{i_{n}}$ the summation being taken over all ordered partitions of r, where the parts are also allowed to be zero. How many terms are there in $h_{r}$ ?

Test yourself ! Improve your mettle in math !
Regards,
Nalin Pithwa.

Pre RMO or PRMO problem set in elementary combinatorics

Posted by Nalin Pithwa

1) How many maps are there from an n-set to an m-set? How many of these are onto? How many are one-one? Under what conditions?

2) Consider the letters of the word DELHI. Let us form new words, whether or not meaningful, using these letters. The *length* of a word is the number of letters in it, e.g., the length of “Delhi” is 5; the length of “Hill” is 4. Answer the following questions when (a) repetition of letters is not allowed and (b) repetition of the letters is allowed:
(i) How many words can be formed of length 1,2,3,4,…?
(ii) How many words in (i) will consist of all the letters?
(iii) How many words of the words in (i) will consist of 1,2,3,4, …specified letters?
(iv) How many of the words in (i) will consist of only 1,2,3,4,…letters?
(v) How many of the words in (i) will be in the alphabetical order of the letters?

3) Repeat Problem 2 with the word MISSISSIPPI.

4) Suppose there are 5 distinct boxes and we want to sort out 1,2,3,…,n objects into these boxes.
4i) In how many ways can this be done?
4ii) In how many of these situations would no box be empty?
4iii) In how many of the above would only 1,2,3,4 … specified boxes be occupied?
4iv) In how many would only 1,2,3,4…boxes be occupied?
4iv) If the objects are indistinguishable from one another, how would the answers to (i) to (iv) change, it at all?

If there is an added restriction that each box can hold only one object and no more, what will be the answers to (i) to (v)?

5) Repeat Problem 4 with 9 boxes.

6) Repeat Problem 4 with 5 non-distinct (=indistinguishable identical) boxes.

7) Repeat Problem 6 with 9 boxes.

8) How many 5-letter words of binary digits are there?

9) Ten teams participate in a tournament. The first team is awarded a gold medal, the second a silver medal, and the third a bronze medal. In how many ways can the medals be distributed?

10) The RBI prints currency notes in denominations of One Rupee, Two Rupees, Five Rupees, Ten Rupees, Twenty Rupees, Fifty Rupees, and One hundred rupees. In how many ways can it display 10 currency notes, not necessarily of different denominations? How many of these will have all denominations?

11) In how many ways can an employer distribute INR 100/- as Holiday Bonus to his 5 employees? No fraction of a rupee is allowed. Also, do not worry about question of equity and fairness!

12) The results of 20 chess games (win, lose, or draw) have to be predicted. How many different forecasts can contain exactly 15 correct results?

13) How many distinct results can we obtain from one throw of four dice? five dice? Can you generalize this?

14) In how many ways can 8 rooks be placed on a standard chess board so that no rook can attack another? How many if the rooks are labelled? How would the answer be modified if we remove the restriction that “no rook can attack another”?

15) Show that there are 7 partitions of the integer 5, and 33 partitions of the integer 9. How many of these have 4 parts ? How many have the largest part equal to 4? Experiment with other partitions and other numbers.

Cheers,
Nalin Pithwa.

Elementary problems in Ramsey number theory for RMO

Posted by Nalin Pithwa

Question 1:

Show that in any group of 6 people there will always be a subgroup of 3 people who are pairwise acquainted or a subgroup of 3 people who are pairwise strangers.

Solution 1:

Let $\{ A, B, C, D, E, F\}$ be a group of 6 people. Suppose that the people known to A are seated in room Y and the people NOT known to A are seated in room Z; A is not in either room. Then, there are necessarily at least 3 people in either room Y or in room Z; (a) Suppose B, C, D to be in room Y. Either these 3 people are mutual strangers (and so the given theorem is true), or, at least two of them (say, B and C) know each other. In the latter case, A, B and C form a group of 3 mutual acquaintances — and again, the theorem is true. (b) In (a), replace room Y by Z and interchange the notion of ‘”acquaintances” and “strangers”‘.

Question 2:

Show that in any group of 10 people there is always (a) a subgroup of 3 mutual strangers or a subgroup of 4 mutual acquaintances, and (b) a subgroup of 3 mutual acquaintances or a subgroup of 4 mutual strangers.

Solution 2:

(a) Let A be one of the ten people; the remaining 9 people can be assigned to two rooms: those who are known to A are in room Y and those who are not known to A are in room Z. Either room Y has at least 6 people or room Z has at least 6 people. For, (i) suppose room Y has at least 6 people. Then, by previous problem number 1, there is either a subgroup of 3 mutual acquaintances or a subgroup of 3 mutual strangers (thus, the theorem is true) in this room. In the former case, A and these 3 people constitute 4 mutual acquaintances (ii) Suppose room Z has at least 4 people. Either these 4 people know one another or at least 2 of them, say B and C, do not know each other. In the former case, we have a subgroup of 4 mutual acquaintances. In the latter case A, B and C constitute 3 mutual strangers.

(b) In the previous scenario, let people who are strangers become acquaintances, and let people who are acquaintances pretend they are strangers. The situation is symmetric.

Question 3:

Show that in any subgroup of 20 people there will always be either a subgroup of 4 mutual acquaintances or a subgroup of 4 mutual strangers.

Solution 3:

Suppose A is one of these 20 people. People known to A are in room Y and people not known to A are room Z. Either room Y has at least 10 people or room Z has at least 10 people. (i) If Y has at least 10 people, then by part B of problem number 2 here, there is either a subgroup of 3 mutual acquaintances or a subgroup of 4 mutual strangers — as asserted — in this room. In the former case, A and these mutual acquaintances will form a subgroup of 4 mutual acquaintances. (ii) Switch ‘”acquaintances” and “strangers”‘ in (i).

Question 4:

Let p and q be 2 positive integers. A positive integer r is said to have the (p,q) – Ramsey property, if in any group of r people either there is a subgroup of p people known to one another or there is a subgroup of q people not known to one another. {By Ramsey’s theorem, all sufficiently large integers r have the (p,q)-Ramsey property.} The smallest r with the (p,q)-Ramsey property is called the Ramsey number R(p,q). Show that (a) R(p,q) = R (q,p). (b) $R(p,1)=1$ , and (c) R(p,2)=p.

Solution 4:

(a) By parts (b) of the previous three questions, we have proved part a of the proof here.

(b) This is obvious.

(c) In any group of p people, if all of them are not known to one another, there will be at least 2 people who do not know each other.

Question 5:

Prove that $R(3,3)=6$ .

Solution 5:

Question 1 and its proof in this blog article imply that $R(3,3) \leq 6$ .

To prove that $R(3,3)>5$ , it is sufficient to consider a seating arrangement of 5 people about a round table in which each person knows only the 2 people on either side. In such a situation, there is no set of 3 mutual acquaintances and no set of 3 people not known to one another.

Question 6:

Show that if m and n are integers both greater than 2, then

$R(m,n) \leq R(m-1,n) + R(m,n-1)$ .

(this recursive inequality gives a non-sharp upper bound for R(m,n)).

Solution 6:

Let $p \equiv R(m-1,n)$ , $q=R(m,n-1)$ and $r \equiv p + q$ . Consider a group $\{ 1,2, 3, \ldots, r\}$ of r people. Let L be the set of people known to person 1 and M be the set of people NOT known to person 1. The two sets together have $r-1$ people, so either L has at least p people or M has at least q people. (a) If L has $p \equiv R(m-1,n)$ people, then, by definition, it contains a subset of $(m-1)$ people known to one another or it contains a subset of n people unknown to one another. In the former case, the $(m-1)$ people and person 1 constitute m people known to one another.

Thus, in their case, a group of $R(m-1,n) + R(m,n-1)$ people necessarily includes m mutual acquaintances or n mutual strangers. That is, $R(m,n) \leq R(m-1,n) + R(m,n-1)$ .

(b) By the usual symmetry argument, the same conclusion follows when M contains q people.

Question 7:

(Remark: A pretty property of Ramsey numbers related to combinatorics).

Show that if m and n are integers greater than 1, then $R(m,n) \leq { {m+n-2} \choose {m-1}}$ — a non-recursive upper bound.

Solution 7:

When $m=2$ , or $n=2$ , (i) holds with equality (see problem 4 in this blog article). The proof is by induction on $k=m+n$ . As we have just seen, the result is true when $k=4$ . Assume the result true for $k-1$ . Then,

$R(m-1,n) \leq {{m+n-3} \choose {m-2}}$ and $R(m,n-1) \leq {{m+n-3} \choose {m-1}}$

Now, Pascal’s identity gives:

${{m+n-3} \choose {m-2}} + {{m+n-3} \choose {m-1}} = {{m+n-2}} \choose {m-1}$ so that $R(m-1,n) + R(m,n-1) \leq {{m+n-2}} \choose {m-1}$

But, from the previous question and its solution, we get $R(m,n) \leq R(m-1,n) + R(m, n-1)$

PS: As Richard Feynman, used to say, you will have to “piddle” with smallish problems as particular cases of these questions in order to get a grip over theory or formal language of this introduction.

PS: Additionally, you can refer to any basic Combinatorics text like Brualdi, or Alan Tucker or even Schaum Series outline ( V K Balakrishnan).

How to find the number of proper divisors of an integer and other cute related questions

Posted by Nalin Pithwa

Question 1:

Find the number of proper divisors of 441000. (A proper divisor of a positive integer n is any divisor other than 1 and n):

Solution 1:

Any integer can be uniquely expressed as the product of powers of prime numbers (Fundamental theorem of arithmetic); thus, $441000 = (2^{3})(3^{2})(5^{3})(7^{2})$ . Any divisor, proper or improper, of the given number must be of the form $(2^{a})(3^{b})(5^{c})(7^{d})$ where $0 \leq a \leq 3$ , $0 \leq b \leq 2$ , $0 \leq c \leq 3$ , and $0 \leq d \leq 2$ . In this paradigm, the exponent a can be chosen in 4 ways, b in 3 ways, c in 4 ways, d in 3 ways. So, by the product rule, the total number of proper divisors will be $(4)(3)(4)(3)-2=142$ .

Question 2:

Count the proper divisors of an integer N whose prime factorization is: $N=p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} p_{3}^{\alpha_{3}}\ldots p_{k}^{\alpha_{k}}$

Solution 2:

By using the same reasoning as in previous question, the number of proper divisors of N is $(\alpha_{1}+1)(\alpha_{2}+1)(\alpha_{3}+1)\ldots (\alpha_{k}+1)-2$ , where we deduct 2 because choosing all the factors means selecting the given number itself, and choosing none of the factors means selecting the trivial divisor 1.

Question 3:

Find the number of ways of factoring 441000 into 2 factors, m and n, such that $m>1, n>1$ , and the GCD of m and n is 1.

Solution 3:

Consider the set $A = {2^{3}, 3^{2}, 5^{3}, 7^{2}}$ associated with the prime factorization of 441000. It is clear that each element of A must appear in the prime factorization of m or in the prime factorization of n, but not in both. Moreover, the 2 prime factorizations must be composed exclusively of elements of A. It follows that the number of relatively prime pairs m, n is equal to the number of ways of partitioning A into 2 unordered nonempty, subsets (unordered as mn and nm mean the same factorization; recall the fundamental theorem of arithmetic).

The possible unordered partitions are the following:

$A = \{ 2^{3}\} + \{ 3^{2}, 5^{3}, 7^{2}\} = \{3^{2}\}+\{ 2^{3}, 5^{3}, 7^{2}\} = \{ 5^{3}\} + \{ 2^{3}, 3^{2}, 7^{2}\} = \{ 7^{2}\}+\{ 2^{3}, 3^{2}, 5^{3}\}$ ,

and $A = \{ 2^{3}, 3^{2}\} + \{ 5^{3}, 7^{2}\}=\{ 2^{3}, 5^{3}\} + \{3^{2}, 7^{2} \} = \{ 2^{3}, 7^{2}\} + \{ 3^{2}, 5^{3}\}$

Hence, the required answer is $4+3=2^{4-1}+1=7$ .

Question 4:

Generalize the above problem by showing that any integer has $2^{k-1}-1$ factorizations into relatively prime pairs m, n ( $m>1, n>1$ ).

Solution 4:

Proof by mathematical induction on k:

For $k=1$ , the result holds trivially.

For $k=2$ , we must prove that a set of k distinct elements, $Z = \{ a_{1}, a_{2}, a_{3}, \ldots, a_{k-1}, a_{k}\}$ has $2^{k-1}-1$ sets. Now, one partition of Z is

$Z = \{ a_{k}\} \bigcup \{ a_{1}, a_{2}, a_{3}, \ldots, a_{k-1}\} \equiv \{ a_{k}\} \bigcup W$

All the remaining partitions may be obtained by first partitioning W into two parts — which, by the induction hypothesis, can be done in $2^{k-2}-1$ ways — and then, including $a_{k}$ in one part or other — which can be done in 2 ways. By the product rule, the number of partitions of Z is therefore

$1 + (2^{k-2})(2)=2^{k-1}-1$ . QED.

Remarks: Question 1 can be done by simply enumerating or breaking it into cases. But, the last generalized problem is a bit difficult without the refined concepts of set theory, as illustrated; and of course, the judicious use of mathematical induction is required in the generalized case.

Cheers,

Nalin Pithwa.

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