Method of undetermined coefficients for PreRMO, PRMO and IITJEE Foundation maths

  1. Find out when the expression x^{3}+px^{2}+qx+r is exactly divisible by x^{2}+ax+b

Solution 1:

Let x^{3}+px^{2}+qx+r=(x^{2}+ax+b)(Ax+B) where A and B are to be determined in terms of p, q, r, a and b. We can assume so because we know from the fundamental theorem of algebra that the if the LHS has to be of degree three in x, the remaining factor in RHS has to be linear in x.

So, expanding out the RHS of above, we get:

x^{3}+px^{2}+qx+r=Ax^{3}+aAx^{2}+bAx+Bx^{2}+Bax+bB

x^{3}+px^{3}+qx+r=Ax^{3}+(aA+B)x^{2}+x(bA+aB)+bB

We are saying that the above is true for all values of x: hence, coefficients of like powers of x on LHS and RHS are same; we equate them and get a system of equations:

A=1

p=aA+B

bA+aB=q

bB=r

Hence, we get p=a+\frac{r}{b} and bp-ba=r or that b(p-a)=r

Also, b+aB=q so that q=b+\frac{ar}{b} which means q-b=\frac{a}{b}r

but \frac{r}{b}=B=p-a and hence, q-b=\frac{a}{b}(p-a)

So, the required conditions are b(p-a)=r and q-b=\frac{a}{b}(p-a).

2) Find the condition that x^{2}+px+q may be a perfect square.

Solution 2:

Let x^{2}+px+q=(Ax+B)^{2} where A and B are to be determined in terms of p and q; finally, we obtain the relationship required between p and q for the above requirement.

x^{2}+px+q=A^{2}x^{2}+B^{2}+2ABx which is true for all real values of x;

Hence, A^{2}=1 so A=1 or A=-1

Also, B^{2}=q and hence, B=\sqrt{q} or B=-\sqrt{q}

Also, 2AB=p so that 2\sqrt{q}=p so q=\frac{p^{2}}{4}, which is the required condition.

3) To prove that x^{4}+px^{3}+qx^{2}+rx+s is a perfect square if (q-\frac{p^{2}}{4})^{2}=4s and r^{2}=p^{2}s.

Proof 3:

Let x^{4}+px^{3}+qx^{2}+rx+s=(Ax^{2}+Bx+C)^{2}

x^{4}+px^{3}+qx^{2}+rx+s=A^{2}x^{4}+B^{2}x^{2}+C^{2}+2ABx^{3}+2BCx+2ACx^{2}

A^{2}=1

2AB=p

q=B^{2}+2AC

2BC=r

C^{2}=s

A=1 or A=-1

2AB=p \longrightarrow 2B=p \longrightarrow B=\frac{p}{2}

q=B^{2}+2AC=\frac{p^{2}}{4}+2\times \sqrt{s} \longrightarrow (q-\frac{p^{2}}{4})^{2}=4s

2 \times \frac{p}{2} \times \sqrt{s}=r \longrightarrow r^{2}=p^{2}s

More later,

Nalin Pithwa.

PS: Note in the method of undetermined coefficients, we create an identity expression which is true for all real values of x.

Miscellaneous Algebra: pRMO, IITJEE foundation maths 2019

For the following tutorial problems, it helps to know/remember/understand/apply the following identities (in addition to all other standard/famous identities you learn in high school maths):

a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)

By the way, I hope you also know how to derive the above.Let me mention two methods to derive the above :

Method I: Using polynomial division in three variable, divide the dividend a^{3}+b^{3}+c^{3}-3abc by the divisor a+b+c.

Method II: Assume that P(X) is a polynomial with roots a, b and c. So, we know by the fundamental theorem of algebra that P(X)=(X-a)(X-b)(X-c). Now, we also know that a, b and c satisfy P(X). Now, proceed further and complete the proof.

Let us now work on the tutorial problems below:

1) If 2s=a+b+c, prove that \frac{1}{s-a} + \frac{1}{s-b} + \frac{1}{s-c} = \frac{abc}{s(s-a)(s-b)(s-c)}

2) If x^{2}+a^{2}=2(xy+yz+zu-y^{2}-z^{2}), prove that x=y=z=u.

Prove the following identities:

3) b(x^{3}+a^{3})+ax(x^{2}-a^{2})+a^{3}(x+a)=(a+b)(x+a)(x^{2}-ax+a^{2})

4) (ax+by)^{2}+(ay-bx)^{2}+c^{2}x^{2}+c^{2}y^{2}=(x^{2}+y^{2})(a^{2}+b^{2}+c^{2})

5) (x+y)^{3}+ 3(x+y)^{2}z+3(x+y)z^{2}+z^{3}=(x+z)^{3}+3(x+z)^{2}y+3(x+z)y^{2}+y^{3}

6) (a+b+c)(ab+bc+ca)-abc=(a+b)(b+c)(c+a)

7) (a+b+c)^{2}-a(b+c-a)-b(a+c-b)-c(a+b-c)=2(a^{2}+b^{2}+c^{2})

8) (x-y)^{3}+(x+y)^{3}+3(x-y)^{2}(x+y)+3(x+y)^{2}(x-y)=8x^{3}

9) x^{2}(y-z)+y^{2}(z-x)+z^{2}(x-y)+(y-z)(z-x)(z-y)=0

10) a^{3}(b-c)+b^{3}(c-a)+c^{3}(a-b)=-(b-c)(c-a)(a-b)(a+b+c)

11) Prove that (b-c)^{3}+(c-a)^{3}+(a-b)^{3}=3(b-c)(c-a)(a-b)

12) If3 2s=a+b+c, prove that (s-a)^{2}+(s-b)^{2}+(s-c)^{2}+s^{2}=a^{2}+b^{2}+c^{2}

13) If 2s=a+b+c, prove that (s-a)^{3}+(s-b)^{3}+(s-c)^{3}+3abc=s^{3}

14) If 2s=a+b+c, prove that 16s(s-a)(s-b)(s-c)=2b^{2}c^{2}+2c^{2}a^{2}+2a^{2}b^{2}-a^{4}-b^{4}-c^{4}

15) If   2s=a+b+c, then prove that  2(s-a)(s-b)(s-c)+a(s-b)(s-c)+b(s-c)(s-a)+c(s-a)(s-b)=abc

16) If a+b+c=0, then prove that (2a-b)^{3}+(2b-c)^{3}+(2c-a)^{3}=3(2a-b)(2b-c)(2c-a)

17) If a+b+c=0, then prove that \frac{a^{2}}{2a^{2}+bc} + \frac{b^{2}}{2b^{2}+ca} + \frac{c^{2}}{2c^{2}+ab} =1

18) Prove that (x+y+z)^{3}+(x+y-z)^{3}+(x-y+z)^{3}+(x-y-z)^{3}=4x(x^{2}+3y^{2}+3z^{2})

19) If a+b+c=0 prove that (s+3a)^{3}-(s-3b)^{3}-(s-3c)^{3}-3(s-3a)(s-3b)(s-3c)=0

20) If X=b+c-2a, Y=c+a-2b, Z=a+b-2c, find the value of X^{2}+Y^{2}+Z^{2}-3XYZ

21) Prove that (a-b)^{2}+(b-c)^{2}+(c-a)^{2}=2(c-b)(c-a)+2(b-a)(b-c)+2(a-b)(a-c)

22) Prove that a^{2}(b^{3}-c^{3})+b^{2}(c^{3}-a^{3})+c^{2}(a^{3}-b^{3})=(a-b)(b-c)(c-a)(ab+bc+ca)=a^{2}(b-c)^{3}+b^{2}(c-a)^{3}+c^{2}(a-b)^{3} = -[a^{2}b^{2}(a-b)+b^{2}c^{2}(b-c)+c^{2}a^{2}(c-a)]

23) if (a+b)^{2}+(b+c)^{2}+(c+a)^{2}=4(ab+bc+cd), prove that a=b=c=d.

24) If x=a+d, y=b+d, z=c+d, prove that x^{2}+y^{2}+z^{2}-yz-zx-xy=a^{2}+b^{2}+c^{2}-bc-ca-ab

25) If a+b+c=3, prove that \frac{1}{b^{2}+c^{2}-a^{2}}+ \frac{1}{c^{2}+a^{2}-b^{2}} + \frac{1}{a^{2}+b^{2}-c^{2}}=0

26) If a+b+c=0, simplify: \frac{b+c}{bc}(b^{2}+c^{2}-a^{2}) + \frac{c+a}{ca} (c^{2}+a^{2}-b^{2})+ \frac{a+b}{ab}(a^{2}+b^{2}-c^{2})

27) Prove that the equation (x-a)^{2}+(y-b)^{2}+(a^{2}+b^{2}-1)(x^{2}+y^{2}-1)=0 is equivalent to the equation (ax+by-1)^{2}+(bx-ay)^{2}=0, hence show that the only possible values of x and y are: \frac{a}{a^{2}+b^{2}}, \frac{b}{a^{2}+b^{2}}

28) If 2(x^{2}+a^{2}-ax)(y^{2}+b^{2}-by)=x^{2}y^{2}+a^{2}b^{2}, prove that (x-a)^{2}(y-b)^{2}+(bx-ay)^{2}=0 and therefore that a=x and y=b are the only possible solutions.

Good luck for the PreRMo August 2019 !!

Regards,

Nalin Pithwa

 

Cyclic expressions, fractions: Pre RMO, PRMO, IITJEE foundation 2019

In order to solve the following tutorial sheet, it helps to solve/understand and then apply the following beautiful cyclic relations or identities:

(Note if these look new to you, then you need to check the truth of all them; if all are v v familiar to you, just go ahead and crack the tutorial sheet below):

Core Identities in Cyclic Expressions:
1) (b-c)+(c-a)+(a-b)=0
2) a(b-c)+b(c-a)+c(a-b)=0
3) a^{2}(b-c)+b^{2}(c-a)+c^{2}(a-b)=-(a-b)(b-c)(c-a)
4) bc(b-c)+ca(c-a)+ab(a-b)=-(a-b)(b-c)(c-a)
5) a(b^{2}-c^{2})+b(c^{2}-a^{2})+c(a^{2}-b^{2})=(a-b)(b-c)(c-a)

Solve or simplify the following:

1) \frac{a}{(a-b)(a-c)} + \frac{b}{(b-c)(b-a)} + \frac{c}{(c-a)(c-b)}
2) \frac{bc}{(a-b)(a-c)} + \frac{ca}{(b-c)(b-a)} + \frac{ab}{(c-a)(c-b)}
3) \frac{a^{2}}{(a-b)(a-c)} + \frac{b^{2}}{(b-c)(b-a)} + \frac{c^{2}}{(c-a)(c-b)}
4) \frac{a^{3}}{(a-b)(a-c)} + \frac{b^{3}}{(b-c)(b-a)} + \frac{c^{3}}{(c-a)(c-b)}
5) \frac{a(b+c)}{(a-b)(c-a)} + \frac{b(a+c)}{(a-b)(b-c)} + \frac{a(a+b)}{(c-a)(b-c)}
6) \frac{1}{a(a-b)(a-c)} + \frac{1}{b(b-c)(b-a)} + \frac{1}{c(c-a)(c-b)}
7) \frac{bc}{a(a^{2}-b^{2})(a^{2}-c^{2})} + \frac{ca}{b(b^{2}-c^{2})(b^{2}-a^{2})} + \frac{ab}{c(c^{2}-a^{2})(c^{2}-b^{2})}
8) \frac{(x-b)(x-c)}{(a-b)(a-c)} + \frac{(x-c)(x-a)}{(b-c)(b-a)} + \frac{(x-a)(x-b)}{(c-a)(c-b)}
9) \frac{bc(a+d)}{(a-b)(a-c)} + \frac{ca(b+d)}{(b-c)(b-a)} + \frac{ab(c+d)}{(c-a)(c-b)}
10) \frac{1}{(a-b)(a-c)(x-a)} + \frac{1}{(b-c)(b-a)(x-b)} + \frac{1}{(c-a)(c-b)(x-c)}
11) \frac{a^{2}}{(a-b)(a-c)(x+a)} + \frac{b^{2}}{(b-c)(b-a)(x+b)} + \frac{c^{2}}{(c-a)(c-b)(x+c)}
12) a^{2}\frac{(a+b)(a+c)}{(a-b)(a-c)} + b^{2}\frac{(b+c)(b+a)}{(b-c)(b-a)} + c^{2}\frac{(c+a)(c+b)}{(c-a)(c-b)}
13) \frac{a^{3}(b-c)+b^{3}(c-a)+c^{3}(a-b)}{(b-c)^{3}+(c-a)^{3}+(a-b)^{3}}
14) \frac{a^{2}(b-c)+b^{2}(c-a)+c^{2}(a-b)+2(c-a)(a-b)(b-c)}{(b-c)^{3}+(c-a)^{3}+(a-b)^{3}}
15) \frac{a^{3}(b-c)+b^{3}(c-a)+c^{3}(a-b)}{a^{2}(b-c)+b^{2}(c-a)+c^{2}(a-b)}
16) \frac{a^{2}(b-c)^{3}+b^{2}(c-a)^{3}+c^{2}(a-b)^{3}}{(a-b)(b-c)(c-a)}
17) \frac{\frac{b-c}{a} + \frac{c-a}{b} + \frac{a-b}{c}}{\frac{1}{a}(\frac{1}{b^{2}}-\frac{1}{c^{2}})+\frac{1}{b}(\frac{1}{c^{2}}-\frac{1}{a^{2}})+\frac{1}{c}(\frac{1}{a^{2}}-\frac{1}{b^{2}})}^
18) \frac{a^{2}(\frac{1}{a^{2}}-\frac{1}{b^{2}})+b^{2}(\frac{1}{a^{2}}-\frac{1}{c^{2}})+c^{2}(\frac{1}{b^{2}}-\frac{1}{a^{2}})}{\frac{1}{bc}(\frac{1}{c}-\frac{1}{b})+\frac{1}{ca}(\frac{1}{a}-\frac{1}{c})+\frac{1}{ab}(\frac{1}{b}-\frac{1}{c})}
19) \frac{a}{(a-b)(a-c)(x-a)} + \frac{b}{(b-c)(b-a)(x-b)} + \frac{c}{(c-a)(c-b)(x-c)}

More later,
Nalin Pithwa

Check your talent: are you ready for math or mathematical sciences or engineering

At the outset, let me put a little sweetener also: All I want to do is draw attention to the importance of symbolic manipulation. If you can solve this tutorial easily or with only a little bit of help, I would strongly feel that you can make a good career in math or applied math or mathematical sciences or engineering.

On the other hand, this tutorial can be useful as a “miscellaneous or logical type of problems” for the ensuing RMO 2019.

I) Let S be a set having an operation * which assigns an element a*b of S for any a,b \in S. Let us assume that the following two rules hold:

i) If a, b are any objects in S, then a*b=a

ii) If a, b are any objects in S, then a*b=b*a

Show that S can have at most one object.

II) Let S be the set of all integers. For a, b in S define * by a*b=a-b. Verify the following:

a) a*b \neq b*a unless a=b.

b) (a*b)*c \neq a*{b*c} in general. Under what conditions on a, b, c is a*(b*c)=(a*b)*c?

c) The integer 0 has the property that a*0=a for every a in S.

d) For a in S, a*a=0

III) Let S consist of two objects \square and \triangle. We define the operation * on S by subjecting \square and \triangle to the following condittions:

i) \square * \triangle=\triangle = \triangle * \square

ii) \square * \square = \square

iii) \triangle * \triangle = \square

Verify by explicit calculation that if a, b, c are any elements of S (that is, a, b and c can be any of \square or \triangle) then:

i) a*b \in S

ii) (a*b)*c = a*(b*c)

iii) a*b=b*a

iv) There is a particular a in S such that a*b=b*a=b for all b in S

v) Given b \in S, then b*b=a, where a is the particular element in (iv) above.

This will be your own self-appraisal !!

Regards,

Nalin Pithwa

Some random problems in algebra (part b) for RMO and INMO training

1) Solve in real numbers the system of equations:

y^{2}+u^{2}+v^{2}+w^{2}=4x-1

x^{2}+u^{2}+v^{2}+w^{2}=4y-1

x^{2}+y^{2}+v^{2}+w^{2}=4u-1

x^{2}+y^{2}+u^{2}+w^{2}=4v-1

x^{2}+y^{2}+u^{2}+v^{2}=4w-1

Hints: do you see some quadratics ? Can we reduce the number of variables? …Try such thinking on your own…

2) Let a_{1}, a_{2}, a_{3}, a_{4}, a_{5} be real numbers such that a_{1}+a_{2}+a_{3}+a_{4}+a_{5}=0 and \max_{1 \leq i <j \leq 5} |a_{i}-a_{j}| \leq 1. Prove that a_{1}^{2}+a_{2}^{2}+a_{3}^{2}+a_{4}^{2}+a_{5}^{2} \leq 10.

3) Let a, b, c be positive real numbers. Prove that

\frac{1}{2a} + \frac{1}{2b} + \frac{1}{2d} \geq \frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{c+a}

More later

Nalin Pithwa.

Some random assorted (part A) problems in algebra for RMO and INMO training

You might want to take a serious shot at each of these. In the first stage of attack, apportion 15 minutes of time for each problem. Do whatever you can, but write down your steps in minute detail. In the last 5 minutes, check why the method or approach does not work. You can even ask — or observe, for example, that if surds are there in an equation, the equation becomes inherently tough. So, as a child we are tempted to think — how to get rid of the surds ?…and so on, thinking in math requires patience and introversion…

So, here are the exercises for your math gym today:

1) Prove that if x, y, z are non-zero real numbers with x+y+z=0, then

\frac{x^{2}+y^{2}}{x+y} + \frac{y^{2}+z^{2}}{y+z} + \frac{z^{2}+x^{2}}{x+z} = \frac{x^{3}}{yz} + \frac{y^{3}}{zx} + \frac{z^{3}}{xy}

2) Let a b, c, d be complex numbers with a+b+c+d=0. Prove that

a^{3}+b^{3}+c^{3}+d^{3}=3(abc+bcd+adb+acd)

3) Let a, b, c, d be integers. Prove that a+b+c+d divides

2(a^{4}+b^{4}+c^{4}+d^{4})-(a^{2}+b^{2}+c^{2}+d^{2})^{2}+8abcd

4) Solve in complex numbers the equation:

(x+1)(x+2)(x+3)^{2}(x+4)(x+5)=360

5) Solve in real numbers the equation:

\sqrt{x} + \sqrt{y} + 2\sqrt{z-2} + \sqrt{u} + \sqrt{v} = x+y+z+u+v

6) Find the real solutions to the equation:

(x+y)^{2}=(x+1)(y-1)

7) Solve the equation:

\sqrt{x + \sqrt{4x + \sqrt{16x + \sqrt{\ldots + \sqrt{4^{n}x+3}}}}} - \sqrt{x}=1

8) Prove that if x, y, z are real numbers such that x^{3}+y^{3}+z^{3} \neq 0, then the ratio \frac{2xyz - (x+y+z)}{x^{3}+y^{3}+z^{3}} equals 2/3 if and only if x+y+z=0.

9) Solve in real numbers the equation:

\sqrt{x_{1}-1} = 2\sqrt{x_{2}-4}+ \ldots + n\sqrt{x_{n}-n^{2}}=\frac{1}{2}(x_{1}+x_{2}+ \ldots + x_{n})

10) Find the real solutions to the system of equations:

\frac{1}{x} + \frac{1}{y} = 9

(\frac{1}{\sqrt[3]{x}} + \frac{1}{\sqrt[3]{y}})(1+\frac{1}{\sqrt[3]{x}})(1+\frac{1}{\sqrt[3]{y}})=18

More later,
Nalin Pithwa

PS: if you want hints, do let me know…but you need to let me know your approach/idea first…else it is spoon-feeding…

A Primer: Generating Functions: Part II: for RMO/INMO 2019

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. For example, let there be only two symbols \alpha_{1}, \alpha_{2}. Let us look for combinations of the form:

\alpha_{1}, \alpha_{2}, \alpha_{1}\alpha_{2}, \alpha_{1}\alpha_{1}, \alpha_{2}\alpha_{2}, \alpha_{1}\alpha_{1}\alpha_{2}, \alpha_{1}\alpha_{2}\alpha_{2}, \alpha_{1}\alpha_{1}\alpha_{2}\alpha_{2}

where, in each combination, each symbol may occur once, twice, or not at all. The OGF for this can be constructed by reasoning as follows: the choices for \alpha_{1} are not-\alpha_{1}, \alpha_{1} once, \alpha_{1} twice. This is represented by the factor (1+\alpha_{1}t+\alpha_{1}^{2}t^{2}). Similarly, the possible choices for \alpha_{2} correspond to the factor (1+\alpha_{2}t+\alpha_{2}^{2}t^{2}). So, the required OGF is (1+\alpha_{1}t+\alpha_{1}^{2}t)(1+\alpha_{2}t+\alpha_{2}^{2}t^{2})

On expansion, this gives : 1+(\alpha_{1}+\alpha_{2})t+(\alpha_{1}\alpha_{2}+\alpha_{1}^{2}+\alpha_{2}^{2})t^{2}+(\alpha_{1}^{2}\alpha_{2}+\alpha_{1}\alpha_{2}^{2})t^{3}+(\alpha_{1}^{2}\alpha_{2}^{2})t^{4}

Note that if we omit the term 1 (which corresponds to not choosing any \alpha), the other 8 terms correspond to the 8 different combinations listed in (*). Also, observe that the exponent r of the t^{r} tells us that the coefficient of t^{r} has the list or inventory of the r-combinations (under the required specification — in this case, with the restriction on repetitions of symbols) in it:

\bf{Illustration}

In the light of the foregoing discussion, let us now take up the following question again: in how many ways, can a total of 16 be obtained by rolling 4 dice once?; the contribution of each die to the total is either a “1” or a “2” or a “3” or a “4” or a “5” or a “6”. The contributions from each of the 4 dice have to be added to get the total — in this case, 16. So, if we write: t^{1}+t^{2}+t^{3}+t^{4}+t^{5}+t^{6}

as the factor corresponding to the first die, the factors corresponding to the other three dice are exactly the same. The product of these factors would be:

(*) (t+t^{2}+t^{3}+t^{4}+t^{5}+t^{6})^{4}

Each term in the expansion of this would be a power of t, and the exponent k of such a term t^{k} is nothing but the total of the four contributions which went into it. The number of times a term t^{k} can be obtained is exactly the number of times k can be obtained as a total on a throw of the four dice. So, if \alpha_{k} is the coefficient of t^{k} in the expansion, \alpha_{16} is the answer for the above question. Further, since (*) simplifies to (\frac{t(1-t^{6})}{1-t})^{4}, it follows that the answer for the above question tallies with the coefficient specified in the following next question: calculate the coefficient of t^{12} in (\frac{(1-t^{6})}{(1-t)})^{4}.6

Now, consider the following problem: Express the number N(n,p) of ways of obtaining a total of n by rolling p dice, as a certain coefficient in a suitable product of binomial expansions in powers of t. [ this in turn, is related to the observation that the number of ways a total of 16 can be obtained by rolling 4 dice once is the same as the coefficient of t^{12} in (\frac{1-t^{6}}{1-t})^{4}]:

So, we get that N(n,p)= coefficient of t^{n-p} in (\frac{1-t^{6}}{1-t})^{p}

Let us take an example from a graphical enumeration:

A \it {graph} G=G(V,F) is a set V of vertices a, b, c, …, together with a set E=V \times V of \it {edges} (a,b), (a,a), (b,a), (c,b), \ldots If (x,y) is considered the same as (y,x), we say the graph is \it{undirected}. Otherwise, the graph is said to be \it{directed}, and we say ‘(a,b) has a direction from a to b’. The edge (x,x) is called a loop. The graph is said to be of order |V|.

If the edge-set E is allowed to be a multiset, that is, if an edge (a,b) is allowed to occur more than once, (and, this may be called a ‘multiple edge’), we refer to the graph as a general graph.

If \phi_{5}(n) and \psi_{5}(n) denote the numbers of undirected (respectively, directed) loopless graphs of order 5, with n edges, none of them a multiple edge, find the series \sum \phi_{5}(n)t^{n} and \sum \psi_{5}(n)t^{n}.

Applying our recently developed techniques to the above question, a graph of 5 specified vertices is uniquely determined once you specify which pairs of vertices are ‘joined’. Suppose we are required to consider only graphs with 4 edges. This would need four pairs of vertices to be selected out of the total of 5 \choose 2 equal to 10 pairs that are available. So selection of pairs of vertices could be made in 10 \choose 4 ways. Each such selection corresponds to one unique graph, with the selected pairs being considered as edges. More informally, having selected a certain pairs of vertices, imagine that the vertices are represented by dots in a diagram and join the vertices of each selected pair by a running line. Then, the “graph” becomes a “visible” object. Note that the number of graphs is just the number of selections of pairs of vertices. Hence, \phi_{5}(4)=10 \choose 4.

Or, one could approach this problem in a different way. Imagine that you have a complete graph on 5 vertices — the “completeness” here means that every possible pair of vertices has been joined by an edge. From the complete graph which has 10 edges, one has to choose 4 edges — any four, for that matter — in order to get a graph as required by the problem.

On the same lines for a directed graph, one has a universe of 10 by 2, that is, 29 edges to choose from, for, each pair x,y gives rise to two possible edges (x,y) and (y,x). Hence,

\psi_{5}(4)=20 \choose 4.

Thus, the counting series for labelled graphs on 5 vertices is 1 + \sum_{p=1}^{10} {10 \choose p}t^{p}
and the counting series for directed labelled graphs on 5 vertices is
1+ \sum_{p=1}^{20}{20 \choose p}t^{p}.

Finally, the OGF for increasing words on an alphabet {a,b,c,d,e} with a<b<c<d<e is

(1+at+a^{2}t^{2}+\ldots)(1+bt+b^{2}t^{2}+\ldots)(1+ct+c^{2}t^{2}+\ldots)\times (1+dt+d^{2}t^{2}+\ldots)(1+et+e^{2}t^{2}+\ldots)

The corresponding OE is (1+t+t^{2}+t^{3}+\ldots)^{5} which is nothing but (1-t)^{-5} (this explains the following problem: Verify that the number of increasing words of length 10 out of the alphabet \{a,b,c,d,e \} with a<b<c<d<e is the coefficient of t^{10} in (1-t)^{-5} ).

We will continue this detailed discussion/exploration in the next article.

Until then aufwiedersehen,
Nalin Pithwa