# Combinatorics for RMO : some basics and examples: homogeneous products of r dimensions

Question:

Find the number of homogeneous products of r dimensions that can be formed out of the n letters a, b, c ….and their powers.

Solution:

By division, or by the binomial theorem, we have:

$\frac{1}{1-ax} = 1 + ax + a^{2}x^{2} + a^{3}x^{3} + \ldots$

$\frac{1}{1-bx} = 1+ bx + b^{2}x^{2} + a^{3}x^{3} + \ldots$

$\frac{1}{1-cx} = 1 + cx + c^{2}x^{2} + c^{3}x^{3} + \ldots$

Hence, by multiplication,

$\frac{1}{1-ax} \times \frac{1}{1-bx} \times \frac{1}{1-cx} \times \ldots$

$= (1+ax + a^{2}x^{2}+a^{3}x^{3}+ \ldots)(1+bx + b^{2}x^{2} + b^{3}x^{3}+ \ldots)(1+cx + c^{2}x^{2} + c^{3}x^{3}+ \ldots)\ldots$

$= 1 + x(a + b + c + \ldots) +x^{2}(a^{2}+ab+ac+b^{2}+bc + c^{2} + \ldots) + \ldots$

$= 1 + S_{1}x + S_{2}x^{2} + S_{3}x^{3} + \ldots$ suppose;

where $S_{1}$, $S_{2}$, $S_{3}$, $\ldots$ are the sums of the homogeneous products of one, two, three, … dimensions that can be formed of a, b, c, …and their powers.

To obtain the number of these products, put a, b, c, …each equal to 1; each term in $S_{1}$, $S_{2}$, $S_{3}$, …now becomes 1, and the values of $S_{1}$, $S_{2}$, $S_{3}$, …so obtained give the number of the homogeneous products of one, two, three, ….dimensions.

Also,

$\frac{1}{1-ax} \times \frac{1}{1-bx} \times \frac{1}{1-cx} \ldots$

becomes $\frac{1}{(1-x)^{n}}$, or $(1-x)^{-n}$

Hence, $S_{r} =$ the coefficient of $x^{r}$ in the expansion of $(1-x)^{-n}$

$= \frac{n(n+1)(n+2)(n+3)\ldots (n+r-1)}{r!}= \frac{(n+r-1)!}{r!(n-1)!}$

Question:

Find the number of terms in the expansion of any multinomial when the index is a positive integer.

In the expansion of $(a_{1}+ a_{2} + a_{3} + \ldots + a_{r})^{n}$

every term is of n dimensions; therefore, the number of terms is the same as the number of homogeneous products of n dimensions that can be formed out of the r quantities $a_{1}$, $a_{2}$, $a_{3}$, …$a_{r}$, and their powers; and therefore by the preceding question and solution, this is equal to

$\frac{(r+n-1)!}{n! (r-1)!}$

A theorem in combinatorics:

From the previous discussion in this blog article, we can deduce a theorem relating to the number of combinations of n things.

Consider n letters a, b, c, d, ….; then, if we were to write down all the homogeneous products of r dimensions, which can be formed of these letters and their powers, every such product would represent one of the combinations, r at a time, of the n letters, when any one of the letters might occur once, twice, thrice, …up to r times.

Therefore, the number of combinations of n things r at a time when repetitions are allowed is equal to the number of homogeneous products of r dimensions which can be formed out of n letters, and therefore equal to $\frac{(n+r-1)!}{r!(n-1)!}$, or ${{n+r-1} \choose r}$.

That is, the number of combinations of n things r at a time when repetitions are allowed is equal to the number of combinations of $n+r-1$ things r at a time when repetitions are NOT allowed.

Example 1:

Find the coefficient of $x^{r}$ in the expansion of $\frac{(1-2x)^{2}}{(1+x)^{3}}$

Solution 1:

The expression $= (1-4x+4x^{2})(1+p_{1}x+p_{2}x^{2}+ \ldots + p_{r}x^{r}+ \ldots)$, suppose.

The coefficients of $x^{r}$ will be obtained by multiplying $p_{r}$, $p_{r-1}$, $p_{r-2}$ by 1, -4, and 4 respectively, and adding the results; hence,

the required coefficient is $p_{r} - 4p_{r-1}+4p_{r-2}$

But, with a little work, we can show that $p_{r} = (-1)^{r}\frac{(r+1)(r+2)}{2}$.

Hence, the required coefficient is

$= (-1)^{r}\frac{(r+1)(r+2)}{2} - 4(-1)^{r-1}\frac{r(r+1)}{2} + 4 (-1)^{r-2}\frac{r(r-1)}{2}$

$= \frac{(-1)^{r}}{2}\times ((r+1)(r+2) + 4r(r+1) + 4r(r-1))$

$= \frac{(-1)^{r}}{2}(9r^{2}+3r+2)$

Example 2:

Find the value of the series

$2 + \frac{5}{(2!).3} + \frac{5.7}{3^{2}.(3!)} + \frac{5.7.9}{3^{3}.(4!)} + \ldots$

Solution 2:

The expression is equal to

$2 + \frac{3.5}{2!}\times \frac{1}{3^{2}} + \frac{3.5.7}{3!}\times \frac{1}{3^{3}} + \frac{3.5.7.9}{4!}\times \frac{1}{3^{4}} + \ldots$

$= 2 + \frac{\frac{3}{2}.\frac{5}{2}}{2!} \times \frac{2^{2}}{3^{2}} + \frac{\frac{3}{2}.\frac{5}{2}.\frac{7}{2}}{3!} \times \frac{2^{3}}{3^{3}} + \frac{\frac{3}{2}.\frac{5}{2}.\frac{7}{2}.\frac{9}{2}}{4!} \times \frac{2^{4}}{3^{4}} + \ldots$

$= 1 + \frac{\frac{3}{2}}{1} \times \frac{2}{3} + \frac{\frac{3}{2}.\frac{5}{2}}{2!} \times (\frac{2}{3})^{2} + \frac{\frac{3}{2}.\frac{5}{2}.\frac{7}{2}}{3!} \times (\frac{2}{3})^{3} + \frac{\frac{3}{2}.\frac{5}{2}.\frac{7}{2}.\frac{9}{2}}{4!} \times (\frac{2}{3})^{4} + \ldots$

$= (1-\frac{2}{3})^{\frac{-3}{2}} = (\frac{1}{3})^{-\frac{3}{2}} = 3^{\frac{3}{2}} = 3 \sqrt{3}$.

Example 3:

If n is any positive integer, show that the integral part of $(3+\sqrt{7})^{n}$ is an odd number.

Solution 3:

Suppose I to denote the integral and f the fractional part of $(3+\sqrt{7})^{n}$.

Then, $I + f = 3^{n} + {n \choose 1}3^{n-1}\sqrt{7} + {n \choose 2}3^{n-2}.7 + {n \choose 3}3^{n-3}.(\sqrt{7})^{3}+ \ldots$…call this relation 1.

Now, $3 - \sqrt{7}$ is positive and less than 1, therefore $(3-\sqrt{7})^{n}$ is a proper fraction; denote it by $f^{'}$;

Hence, $f^{'} = 3^{n} - {n \choose 1}.3^{n-1}.\sqrt{7} + {n \choose 2}.3^{n-2}.7 - {n \choose 3}.3^{n-3}.(\sqrt{7})^{3}+ \ldots$…call this as relation 2.

Add together relations 1 and 2; the irrational terms disappear, and we have

$I + f + f^{'} = 2(3^{n} + {n \choose 2}.3^{n-2}.7+ \ldots ) = an even integer$

But, since f and $f^{'}$ are proper fractions their sum must be 1;

Hence, I is an odd integer.

Nalin Pithwa.

# Science Lives: Laslo Lovasz: Discrete Maths, Combinatorics and Computer Science

Thanks to Simon Foundation : a youtube video.

# Inequalities and mathematical induction: RMO sample problems-solutions

Problem:

1. Prove the inequality —- $2^{n}(n!)^{2} \leq (2n)!$ for all natural numbers greater than or equal to 1.

Proof 1:

First consider the following: $2.6. 10.14 \ldots (4n-2)=\frac{(2n)!}{n!}$. Let us prove this claim first and then use it to prove what is asked: Towards, that end, consider

$RHS = \frac{(2n)(2n-1)(2n-2)(2n-3)(2n-4)\ldots 4.2.1}{1.2.3.4\ldots (n-1)n}=LHS$, cancelling off the common factors in numerator and denominator of RHS. (note this can also be proved by mathematical induction! 🙂 )

In the given inequality:

we need to prove $2^{n}(n!)^{2} \leq (2n)!$

consider $2.6.10.14. \ldots (4n-2)=\frac{(2n)!}{n!}$ where

LHS = $(2.1)(2.3) (2.5) (2.7) \ldots 2(n-1)$, this is an AP with first term 2 and nth term $(4n-2)$ and common difference 4; there are n factors “2”; hence, we $2^{n}1.3.5.7 \ldots (n-1)=\frac{(2n)!}{n!}$ so we get

$2^{n} (n!) 1.3.5.7.\ldots (n-1) = (2n)!$; multiplying and dividing RHS of this by $2.4.6.8.\ldots n$, we get the desired inequality. Remember the inequality is less than or equal to.

Problem 2:

Establish the Bernoulli inequality: If $(1+a) > 0$, then $(1+a)^{n} \geq 1+na$.

Solution 2:

Apply the binomial theorem, which in turn, is proved by mathematical induction ! 🙂

Problem 3:

For all natural numbers greater than or equal to 1, prove the following by mathematical induction:

$\frac{1}{1^{2}} + \frac{1}{2^{2}} + \frac{1}{3^{2}} + \ldots + \frac{1}{n^{2}} \leq 2-\frac{1}{n}$

Proof 3:

Let the given proposition be P(n).

Step 1: Check if P(1) is true. Towards that end:

$LHS=\frac{1}{1^{2}}=1$ and $latex RHS=2-\frac{1}{1}=2-1=1$ and hence, P(1) is true.

Step 2: Let P(n) be true for some $n=k$, $k \in N$. That is, the following is true:

$\frac{1}{1^{2}} + \frac{1}{2^{2}} + \frac{1}{3^{2}} + \ldots + \frac{1}{k^{2}} \leq 2 -\frac{1}{k}$

Add $\frac{1}{(k+1)^{2}}$ to both sides of above inequality, we get the following:

$\frac{1}{1^{2}} + \frac{1}{2^{2}} + \frac{1}{3^{2}} + \ldots + \frac{1}{k^{2}} + \frac{1}{(k+1)^{2}} \leq 2-\frac{1}{k}+\frac{1}{(k+1)^{2}}$

Now, the RHS in above is $2-\frac{1}{k} +\frac{1}{(k+1)^{2}}=2-\frac{k^{2}+k+1}{k(k+1)^{2}}$. We want this to be less than or equal to $2-\frac{1}{k+1}$. Now, $k \in N$, $k>1$, so what we have to prove is the following:

$-\frac{k^{2}+k+1}{k(k+1)^{2}} \leq -\frac{1}{k+1}$, that is, we want to prove that

$(k+1)(k^{2}+k+1) \geq k(k^{2}+2k+1)$, that is, we want $k^{3}+k^{2}+k+k^{2}+k+1 \geq k^{3}+2k^{2}+k$, that is, we want $k+1 \geq 0$, which is obviously true. QED.

Cheers,

Nalin Pithwa.

# RMO Training: more help from Nordic mathematical contest

Problem:

32 competitors participate in a tournament. No two of them are equal and in a one against one match the better always wins. (No tie please). Show that the gold, silver and bronze medal can be found in 39 matches.

Solution:

We begin by determining the gold medallist using classical elimination, where we organize 16 pairs and matches, then 8 matches of the winners, 4 matches of the winners in the second round, then 2-semifinal matches and finally one match making 31 matches altogether.

Now, the second best player must have at some point lost to the best player, and as there were 5 rounds in the elimination, there are 5 candidates for the silver medal. Let $C_{i}$ be the candidate who  lost to the gold medalist in round i. Now, let $C_{1}$ and $C_{2}$ play, the winner play against $C_{3}$, and so forth. After 4 matches, we know the silver medalist; assume this was $C_{k}$.

Now, the third best player must have lost against the gold medalist or against $C_{k}$ or both. (If the player had lost to someone else, there would be at least three better players.) Now, $C_{k}$ won k-1 times in the elimination rounds, the 5-k players $C_{k+1}\ldots C_{5}$ and if k is greater than one, one player $C_{j}$ with $j. So there are either $(k-1)+(5-k)=4$ or $(k-1)+(5-k)+1=5$ candidates for the third place. At most 4 matches are again needed to determine the bronze winners.

Cheers to Norway mathematicians!

Nalin Pithwa.

Reference: Nordic mathematical contests, 1987-2009.

https://www.amazon.in/Nordic-Mathematical-Contest-1987-2009-Todev/dp/1450519830/ref=sr_1_1?s=books&ie=UTF8&qid=1518386661&sr=1-1&keywords=Nordic+mathematical+contest

# RMO 2017 Warm-up: Two counting conundrums

Problem 1:

There are n points in a circle, all joined with line segments. Assume that no three (or more) segments intersect in the same point. How many regions inside the circle are formed in this way?

Problem 2:

Do there exist 10,000 10-digit numbers divisible by 7, all of which can be obtained from one another by a re-ordering of their digits?

Solutions will be put up in a couple of days.

Nalin Pithwa.

# Math concept(s) : simple yet subtle

Most math concepts are intuitive, simple, yet subtle. A similar opinion is expressed by Prof. Michael Spivak in his magnum opus, Differential Geometry (preface). It also reminds me — a famous quote of the ever-quotable Albert Einstein: “everything should be as simple as possible, and not simpler.”

I have an illustrative example of this opinion(s) here:

Consider the principle of mathematical induction:

Most students use the first version of it quite mechanically. But, is it really so? You can think about the following simple intuitive argument which when formalized becomes the principle of mathematical induction:

Theorem: First Principle of Finite Induction:

Let S be a set of positive integers with the following properties:

1. The integer 1 belongs to S.
2. Whenever the integer k is in S, the next integer $k+1$ must also be in S.

Then, S is the set of all positive integers.

The proof of condition 1 is called basis step for the induction. The proof of 2 is called the induction step. The assumptions made in carrying out the induction step are known as induction hypotheses. The induction situation has been likened to an infinite row of dominoes all standing on edge and arranged in such a way that when one falls it knocks down the next in line. If either no domino is pushed over (that is, there is no basis for the induction), or if the spacing is too large (that is, the induction step fails), then the complete line will not fall.

So, also remember that the validity of the induction step does not necessarily depend on the truth of the statement that one is endeavouring to prove.

More later,

Nalin Pithwa.

# Learn hardcore math and programming through games :-) The Game of Nim

Reference: Introductory Combinatorics, fourth edition, Richard A. Brualdi

Example : The Game of Nim:

It is a return to the roots of combinatorics —- which does lie in recreational mathematics and we investigate the ancient game of Nim. (Nim derives from the German word Nimml, meaning Take!) Its solution depends on parity, an important problem-solving concept in combinatorics. We can use parity argument in investigations of perfect covers of chessboards when we show that a board has to have an even number of squares in order that it have a perfect cover with dominoes.

Nim is a game played by two players with heaps of coins (or stones or beans or gems or diamonds !!! 🙂 🙂 …).. Suppose that there are $k \geq 1$ heaps of coins that contain, respectively, $n_{1}$, $n_{2}$, $\ldots$, $n_{k}$ coins. The object of the game is to select the last coin. The rules of the game are the following:

i) The players alternate turns (let us call the player who makes the first move I and then call the other player II).

ii) Each player, when it is his or her turn, selects one of the heaps and removes at least one of the coins from the selected heap. (The player may take all of the coins from the selected heap, thereby leaving an empty heap, which is now “out of play.”)

The game ends when all the heaps are empty. The last player to make a move, that is, the player who takes the last coin(s), is the winner.

The variables in this game are the number k of heaps and the numbers $n_{1}$, $n_{2}$, $\ldots$, $n_{k}$ of coins in the heaps. The combinatorial problem is to determine whether the first or second player wins and how that player should move in order to guarantee a win — a winning strategy.

In order to develop some understanding of Nim, we consider some special cases. (This is an important principle to follow in general: Consider small or special cases in order to develop understanding and intuition. Then try to extend your ideas in order to solve the problem in general.) If there is initially only one heap, then player 1 wins by removing all its coins. Now, suppose that there are $k=2$ heaps, with $n_{1}$, and $n_{2}$ coins, respectively. Whether or not player I can win depends not on the actual values of $n_{1}$ and $n_{2}$ but on whether or not they are equal. Suppose that $n_{1} \neq n_{2}$. Player I can remove enough coins from the larger heap in order to leave two heaps of equal size for player II. Now player I, when it is her turn, can mimic player II’s moves. Thus, if player II takes c coins from one of the heaps, then player I takes the same number c of coins from the other heap. Such a strategy guarantees a win for player I. If $n_{1}=n_{2}$, then player II can win by mimicking player I’s moves. Thus, we have completely solved 2-heap Nim. An example of play in the 2-heap game of Nim with heaps of sizes 8 and 5, respectively, is

$8, 5 \stackrel{I}\rightarrow 5,5 \stackrel{II}\rightarrow 5,2 \stackrel{I}\rightarrow 2,2 \stackrel{II}\rightarrow 0,2 \stackrel{I}\rightarrow 0,0$

The preceding idea in solving 2-heap Nim, namely, moving in such a way as to leave two equal heaps, can be generalized to any number k of heaps. The insight one needs is provided by the concept of the base 2 numeral of an integer. Recall that each positive integer n can be expressed as a base 2 numeral by repeatedly removing the largest power of 2 which does not exceed the number. For instance, to express the decimal number 57 in base 2, we observe that

$2^{6} \leq 57 < 2^{6}$, $57-2^{5}=25$,

$2^{4} \leq 25 < 2^{5}$, $25-2^{4}=9$,

$2^{3} \leq 9 < 2^{4}$, $9-2^{3}=1$,

$2^{0}\leq 1 <2^{1}$, $1-2^{0}=0$.

Thus, $57=2^{5}+2^{4}+2^{3}+2^{0}$,

and the base 2 numeral for 57 is 111001.

Each digit in a base 2 numeral is either 0 or 1. The digit in the ith position, the one corresponding to $2^{i}$, is called the ith bit ($i \geq 0$). We can think of each heap of coins as consisting of subheaps of powers of 2, according to its base numeral. Thus, a heap of size 53 consists of subheaps of sizes $2^{5}$, $2^{4}$, $2^{2}$, and $2^{0}$. In the case of 2-heap Nim, the total number of subheaps of each size is either 0, 1, or 2. There is exactly one subheap of a particular size if and only if the two subheaps of each size is even if and only if the two heaps have the same size — that is, if and only if, player II can win the Nim game.

Now, consider a general Nim game with heaps of sizes $n_{1}$, $n_{2}$, $\ldots$, $n_{k}$. Express each of the numbers $n_{i}$ as base 2 numerals:

$n_{1}=a_{s}\ldots a_{1}a_{0}$

$n_{2}=b_{s}\ldots b_{1}b_{0}$

$\ldots$

$n_{k}=e_{s}\ldots e_{1}e_{0}$

(By including leading 0’s we can assume that all of the heap sizes have base 2 numerals with the same number of digits.) We call a Nim game balanced, provided that the number of subheaps of each size is even. Thus, a Nim game is balanced if and only if:

$a_{s}+b_{s}+\ldots + e_{s}$ is even,

$\vdots$

$a_{i}+b_{i}+\ldots + e_{i}$ is even,

$\vdots$

$a_{0}+b_{0}+\ldots + e_{0}$ is even,

A Nim game that is not balanced is called unbalanced. We say that the ith bit is balanced, provided that the sum $a_{i}+b_{i}+\ldots + e_{i}$ is even, and is unbalanced otherwise. Thus, a balanced game is one in which all bits are balanced, while an unbalanced game is one in which there is at least one unbalanced bit.

We then have the following:

Player I can win in unbalanced Nim games, and player II can win in balanced Nim games.

To see this, we generalize the strategies used in 2-pile Nim. Suppose the Nim game is unbalanced. Let the largest unbalanced bit be the jth bit. Then, player I moves in such a way as to leave a balanced game for player II. She does this by selecting a heap whose jth bit is 1 and removing a number of coins from it so that the resulting game is balanced. No matter what player II does, she leaves for player I an unbalanced game again, and player I once again balances it. Continuing like this ensures player I a win. If the game starts out balanced, then player I’s first move unbalances it, and now player II adopts the strategy of balancing the game whenever it is her move.

For example, consider a 4-pile Nim game with heaps of sizes 7,9, 12 and 15. The base 2 numerals for these heap sizes are, respectively, 0111, 1001, 1100, and 1111. In terms of subheaps of powers of 2 we have:

$\begin{array}{ccccc} \emph{4-pile Nim game} & 2^{3}=8 & 2^{2}=4 & 2^{1}=2 & 2^{0}=1 \\ \emph{heap of size 7} & 0 & 1 & 1 & 1 \\ \emph{heap of size 9} & 1 & 0 & 0 & 1 \\ \emph{heap of size 12} & 1 & 1 & 0 & 0 \\ \emph{heap of size 15} & 1 & 1 & 1 & 1 \\ \end{array}$

This game is unbalanced with the 3rd, 2nd, and 0th bits unbalanced. Player I can remove 11 coins from the pile of size 12, leaving 1 coin. Since the base 2 numeral of 1 is 0001, the game is now balanced. Alternatively, player I can remove 5 coins from the pile of size 9, leaving 4 coins, or player I can remove 13 coins from the pile of size 15, leaving 2 coins.

A quiz:

1. Consider 3-heap Nim with piles of sizes 1, 2 and 4. Show that this game is unbalanced and determine a first move for player I.
2. Is 4-pile Nim with heaps of sizes 22, 19, 14, and 11 balanced or unbalanced? Player I’s first move is to remove 6 coins from the heap of size 19. What should player II’s first move be?
3. Consider 5-pile Nim with heaps of sizes 10, 20, 30, 40, and 50. Is this game balanced? Determine a first move for player I.

Have fun with math games  !

Cheers,

Nalin Pithwa.