Probability Theory Primer: part I

Illustrative Example 1:

What is the chance of throwing a number greater than 4 with a fair die whose faces are numbered from 1 to 6?

Solution 1:

There are 6 possible ways in which the die can fall, and of these two are the favourable events required.

Hence, required chance is \frac{2}{6}=\frac{1}{3}.

Illustrative example 2:

From a bag containing 4 white and 5 black balls a man draws 3 at random; what are the odds against these being all black?

Solution 2:

The total number of ways in which 3 balls can be drawn is 9 \choose 3, and the number of ways of drawing 3 black balls is 5 \choose 3; therefore, the chance of drawing 3 black balls is equal to

\frac{{5 \choose 3}}{{9 \choose 3}}=\frac{5.4.3}{9.3.7}=\frac{5}{43}.

Thus, the odds against the event are 37 to 5.

Illustrative example 3:

Find the chance of throwing at least one ace in a single throw with two dice.

Solution 3:

In die games, there is no ace. We can pick up any number as an ace in this question. Once it is chosen, it is fixed.

So, the possible number of cases is 6 \times 6=36.

An ace on one die may be associated with any of the six numbers on the other die, and the remaining five numbers on the first die may each be associated with the ace on the second die; thus, the number of favourable cases is 11.

Therefore, the required probability is \frac{11}{36}.

Illustrative example 4:

Find the chance of throwing more than 15 in one throw with 3 dice.

Solution 4:

A throw amounting to 18 must be made up of 6, 6, 6 and this can occur in one way only; 17 can be made up of 6. 6. 5 which can occur in 3 ways; 16 may be made up of 6, 6, 4 and 6, 5, 5 each of which arrangements can occur in 3 ways.

Therefore, the number of favourable cases is 1+3+3+3, that is, 10.

And, the total number of cases possible is 6^{3}, that is, 216.

Hence, the required probability is \frac{10}{216}=5/108.

Illustrative example 5:

A has 3 shares in a lottery, in which there are 3 prizes and 6 blanks; B has 1 share in a lottery in which there is 1 prize and 3 blanks; show that A’s chance of success is to B’s as 16:7.

Solution 5:

A may draw 3 prizes in one way; A may draw 2 prizes and 1 blank in \frac{3.3}{1.2} \times 6 ways; A may draw 1 prize and 2 blanks in 3 \times \frac{6.5}{1.2} ways; the sum of these numbers is 64, which is the number of ways in which A can win a prize. Also, he can draw 3 tickets in \frac{9.8.7}{1.2.3} ways, that is, 84 ways.

Hence, A’s chance of success is \frac{64}{84} = \frac{16}{21}.

B’s chance of success is clearly \frac{1}{3}.

Therefore, the required ratio is \frac{\frac{16}{21}}{\frac{1}{3}} = \frac{16}{7}.

Tutorial questions:

  1. In a single throw with two dice, find the chances of throwing (a) a five (b) a six.
  2. From a pack of 52 cards, two cards are drawn at random, find the chance that one is a knave and other is a queen.
  3. A bag contains 5 white, 7 black and 4 red balls. Find the chance that 3 balls all drawn at random are all white.
  4. If four coins are tossed, find the chance that there are two heads and two tails.
  5. One of two events must happen; given that the chance of one is two-thirds that of the other, find the odds in favour of the other.
  6. If from a pack four cards are drawn, find the chance that they will be the four honours of the same suit.
  7. Thirteen persons take their places at a round table, show that it is five to one against two particular persons sitting together.
  8. There are three events A, B, C, out of which one must, and only one can happen; the odds are 8 is to 3 against A; 5 to 2 against B; find the odds against C.
  9. Compare the chance of throwing 4 with one die, 8 with two dice and 12 with three dice.
  10. In shuffling a pack of cards, four are accidentally dropped; find the chance that the missing cards should be one from each suit.
  11. A has 3 shares in a lottery containing 3 prizes and 9 blanks; B has 2 shares in a lottery containing 2 prizes and 6 blanks; compare their chances of success.
  12. Show that the chances of throwing six with 4, 3 or 2 dice respectively are as 1:6:16
  13. There are three works, one consisting of three volumes, one of four volumens, and the other of one volume. They are placed on a shelf at random; prove that the chance that the volumes of the same works are all together is \frac{3}{140}.
  14. A and B throw with two dice; if A throws 9, find B’s chance of throwing a higher number.
  15. The letters forming the word Clifton are placed at random in a row; what is the chance that the two vowels come together?
  16. In a hand, what is the chance that the four kings are held by a specified player?


Nalin Pithwa.






Compilation of elementary results related to permutations and combinations: Pre RMO, RMO, IITJEE math

1. Disjunctive or Sum Rule:

If an event can occur in m ways and another event can occur in n ways and if these two events cannot occur simultaneously, then one of the two events can occur in m+n ways. More generally, if E_{i} (i=1,2,\ldots,k) are k events such that no two of them can occur at the same time, and if E_{i} can occur in n_{i} ways, then one of the k events can occur in n_{1}+n_{2}+\ldots+n_{k} ways.

2. Sequential or Product Rule:

If an event can occur in m ways and a second event can occur in n ways, and if the number of ways the second event occurs does not depend upon how the first event occurs, then the two events can occur simultaneously in mn ways. More generally, $if E_{1} can occur in n_{1}, E_{2} can occur in n_{2} ways (no matter how E_{1} occurs), E_{3} can occur in n_{3} ways (no matter how E_{1} and E_{2} occur), \ldots, E_{k} can occur in n_{k} ways (no matter how the previous k-1 events occur), then the k events can occur simultaneously in n_{1}n_{2}n_{3}\ldots n_{k} ways.

)3. Definitions and some basic relations:

Suppose X is a collection of n distinct objects and r is a nonnegative integer less than or equal to n. An r-permutation of X is a selection of r out of the n objects but the selections are ordered. 

An n-permutation of X is called a simply a permutation of X.

The number of r-permutations of a collection of n distinct objects is denoted by P(n,r); this number is evaluated as follows: A member of X can be chosen to occupy the first of the r positions in n ways. After that, an object from the remaining collections of (n-1) objects can be chosen to occupy the second position in (n-1) ways. Notice that the number of ways of placing the second object does not depend upon how the first object was placed or chosen. Thus, by the product rule, the first two positions can be filled in n(n-1) ways,….and all r positions can be filled in

P(n,r) = n(n-1)\ldots (n-r+1) = \frac{n!}{(n-r)}! ways.

In particular, P(n,n) = n!

Note: An unordered selection of r out of the n elements of X is called an r-combination of X. In other words, any subset of X with r elements is an r-combination of X. The number of r-combinations or r-subsets of a set of n distinct objects is denoted by n \choose r (read as ” n ‘choose’ r). For each r-subset of X there is a unique complementary (n-r)-subset, whence the important relation {n \choose r} = n \choose {n-r}.

To evaluate n \choose r, note that an r-permutation of an n-set X is necessarily a permutation of some r-subset of X. Moreover, distinct r-subsets generate r-permutations each. Hence, by the sum rule:

P(n,r)=P(r,r)+P(r,r)+\ldots + P(r,r)

The number of terms on the right is the number of r-subsets of X. That is, n \choose r. Thus, P(n,r)=P(r,r) \times {n \choose r}=r! \times {n \choose r}.

The following is our summary:

  1. P(n,r) = \frac{n!}{(n-r)!}
  2. {n \choose r}=\frac{P(n,r)}{r!}=\frac{n!}{r! (n-r)!}=n \choose {n-r}

4. The Pigeonhole Principle: Basic Version:

If n pigeonholes (or mailboxes) shelter n+1 or more pigeons (or letters), at least 1 pigeonhole (or mailbox) shelters at least 2 pigeons (or letters).

5. The number of ways in which m+n things can be divided into two groups containing m and n equal things respectively is given by : \frac{(m+n)!}{m!n!}

Note: If m=n, the groups are equal (and hence, indistinguishable), and in this case the number of different ways of subdivision is \frac{(2m)!}{2!m!m!}

6. The number of ways in which m+n+p things can be divided into three groups containing m, n, p things severally is given by: \frac{(m+n+p)!}{m!n!p!}

Note: If we put m=n=p, we obtain \frac{(3m)!}{m!m!m !} but this formula regards as different all the possible orders in which the three groups can occur in any one mode of subdivision. And, since there are 3! such orders corresponding to each mode of subdivision, the number of different ways in which subdivision into three equal groups can be made in \frac{(3m)!}{m!m!m!3!} ways.

7. The number of ways in which n things can be arranged amongst themselves, taking them all at a time, when p of the things are exactly alike of one kind, q of them are exactly alike of a another kind, r of them are exactly alike of a third kind, and the rest are all different is as follows: \frac{n!}{p!q!r!}

8. The number of permutations of n things r at a time, when such things may be repeated once, twice, thrice…up to r times in any arrangement is given by: n^{r}. Cute quiz: In how many ways, can 5 prizes be given away to 4 boys, when each boy is eligible for all the prizes? (Compare your answers with your friends’ answers :-))

9. The total number of ways in which it is possible to make a selection by taking some or all of n things is given by : 2^{n}-1

10. The total number of ways in which it is possible to make a selection by taking some or all out of p+q+r+\ldots things, whereof p are alike of one kind, q alike of a second kind, r alike of a third kind, and so on is given by : (p+1)(q+1)(r+1)\ldots-1.


Nalin Pithwa.

III. Tutorial problems. Symmetric and alternating functions. RMO and IITJEE math

  1. Simplify: (b^{-1}+c^{1})(b+c-a)+(c^{-1}+a^{-1})(c+a-b)+(a^{-1}+b^{-1})(a+b=c)
  2. Simplify: \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)}
  3. Simplify: \frac{b^{2}+c^{2}-a^{2}}{(a-b)(a-c)} + \frac{c^{2}+a^{2}-b^{2}}{(b-c)(b-a)} + \frac{a^{2}+b^{2}-c^{2}}{(c-a)(c-b)}
  4. Simplify: \frac{b-c}{1+bc} + \frac{c-a}{1+ca} + \frac{a-b}{1+ab}
  5. Simplify: \frac{a(b-c)}{1+bc} + \frac{b(c-a)}{1+ca} + \frac{c(a-b)}{1+ab}
  6. Factorize: (b-c)^{2}(b+c-2a)+(c-a)^{2}(c+a-2b)+(a-b)^{2}(a+b-2c). Put b-c=x, c-a=y, a-b=zand b+c-2a=y-z
  7. Factorize: 8(a+b+c)^{2}-(b+c)^{2}-(c+a)^{2}-(a+b)^{2}. Put b+c=x, c+a=y, a+b=z.
  8. Factorize: (a+b+c)^{2}-(b+c-a)^{2}-(c+a-b)^{2}+(a+b-c)^{2}
  9. Factorize: (1-a^{2})(1-b^{2})(1-c^{2})+(a-bc)(b-ac)(c-ab)
  10. Express the following substitutions as the product of transpositions: (i) \left(\begin{array}{cccccc}123456\\654321\end{array}\right) (ii) \left(\begin{array}{cccccc}123456\\246135\end{array}\right) (iii) \left(\begin{array}{cccccc}123456\\641235\end{array}\right)


Nalin Pithwa.


II. tutorial problems. Symmetric and alternating functions. RMO, IITJEE math

Reference: Higher Algebra by Bernard and Child. 

Exercises: (based on the earlier blogged chapter from the above reference):

Prove the identities from problem 1 to 5 given below where \Sigma{\alpha}, \Sigma{\alpha\beta} etc. denote symmetric functions of \alpha, \beta, \gamma, \delta. Also verify by putting \alpha=\beta=\gamma=\delta=1:

1 (\alpha+\beta+\gamma+\delta)(\alpha^{2}+\beta^{2}+\gamma^{2}+\delta^{2}) = \Sigma{\alpha^{2}}+\Sigma{\alpha^{2}\beta}

2. (\alpha+\beta+\gamma+\delta)(\beta\gamma\delta+\gamma\delta\alpha+\delta\alpha\beta+\alpha\beta\gamma) = \Sigma{\alpha^{2}\beta\gamma}+4\alpha\beta\gamma\delta

3. (\beta\gamma\delta+\gamma\delta\alpha+\delta\alpha\beta+\alpha\beta\gamma)^{2}=\Sigma{\alpha^{2}}{\beta^{2}}{\gamma^{2}}+2\Sigma{\alpha\beta\gamma\delta}\Sigma{\alpha\beta}


5. \Sigma{\alpha\beta}.\Sigma{\alpha\beta\gamma}=\Sigma{\alpha^{2}\beta^{2}\gamma}+3\alpha\beta\gamma\delta.\Sigma{\alpha}


Nalin Pithwa.

Tutorial problems. I. Symmetric and Alternating functions. RMO/IITJEE Math


  1. Show that (bc-ad)(ca-bd)(ab-cd) is symmetric with respect to a, b, c, d.
  2. Show that the following expressions are cyclic with respect to a, b, c, d, taken in this order: (a-b+c-d)^{2} and (a-b)(c-d)+(b-c)(d-a)
  3. Expand the expression using \Sigma notation: (y+z-2x)(z+x-2y)(x+y-2z)
  4. Expand the expression using \Sigma notation: (x+y+z)^{2}+(y+z-x)^{2}+(z+x-y)^{2}+(x+y-z)^{2}
  5. Prove that (\beta^{2}\gamma^{2}+\gamma^{2}\alpha^{2}+\alpha^{2}\beta^{2})(\alpha+\beta+\gamma)= \Sigma\alpha^{2}\beta^{2}+\alpha\beta\gamma\Sigma\alpha\beta
  6. Prove that (\alpha-\beta)(\alpha-\gamma)+(\beta-\gamma)(\beta-\alpha)+(\gamma-\alpha)(\gamma-\beta)=\Sigma{\alpha^{2}}-\Sigma{\alpha}{\beta}
  7. Prove that (\beta-\gamma)(\beta+\gamma-\alpha)+(\gamma-\alpha)(\gamma+\alpha-\beta)+(\alpha-\beta)(\alpha+\beta-\gamma)=0
  8. Prove that : \alpha(\beta-\gamma)^{2}+\beta(\gamma-\alpha)^{2}+\gamma(\alpha-\beta)^{2}=\Sigma{\alpha^{2}}{\beta}-6\alpha\beta\gamma
  9. Prove that: (\beta^{2}\gamma+\beta\gamma^{2}+\gamma^{2}\alpha+\gamma\alpha^{2}+\alpha^{2}\beta+\alpha\beta^{2})(\alpha+\beta+\gamma)=\Sigma{\alpha^{2}}\beta+2\Sigma{\alpha^{2}}{\beta^{2}}+2\alpha\beta\gamma\Sigma{\alpha}
  10. Prove that : a^{2}(b+c)+b^{2}(c+a)+c^{2}(a+b)+abc(a+b+c)=\Sigma{a^{2}}.\Sigma{ab}.
  11. Prove that: (a+b-c)(a^{2}+b^{2}-c^{2})+(b+c-a)(b^{2}+c^{2}-a^{2})+(c+a-b)(c^{2}+a^{2}-b^{2})=3\Sigma{a^{3}}-\Sigma{a^{2}{b}}
  12. Prove that: (a^{2}+b^{2}+c^{2})(x^{2}+y^{2}+z^{2})=(ax+by+cz)^{2}+(bz-cy)^{2}+(cx-az)^{2}+(ay-bz)^{2}
  13. Prove that: (b^{2}-ac)(c^{2}-ab)+(c^{2}-ab)(a^{2}-bc)+(a^{2}-bc)(b^{2}-ac)=-(bc+ca+ab)(a^{2}+b^{2}+c^{2}-bc-ca-ab)
  14. Prove that: (a^{2}-bc)(b^{2}-ac)(c^{2}-ab)=abc(a^{2}+b^{2}+c^{2})-(b^{2}c^{2}+c^{2}a^{2}+a^{2}b^{2})
  15. If one of the numbers a, b, and c is the geometric mean of the other two, use the previous problem to prove the following: abc(a^{2}+b^{2}+c^{2})=b^{2}c^{2}+c^{2}a^{2}+a^{2}b^{2}
  16. If the numbers x, y, z taken in some order or other form an AP, use problem 3 to prove that 2(x+y+z)^{2}+27xyz=9(x+y+z)(yz+zx+xy)
  17. Express 2(a-b)(a-c)+2(b-c)(b-a)+2(c-a)(c-b) as the sum of three squares. Hence, show that (b-c)(c-a)+(c-a)(a-b)+(a-b)(b-c) is negative for all real values of a, b, c except when a=b=c. Hint: Put b-c=x, c-a=y, a-b=z, and notice that x^{2}+y^{2}+z^{2}+2(xy+yz+zx)=(x+y+z)^{2}=0.
  18. If x+y+z=0, show that (i) 2yz=x^{2}-y^{2}-z^{2}; (ii) (y^{2}+z^{2}-x^{2})(z^{2}+z^{2}-y^{2})(x^{2}+y^{2}-z^{2})+8x^{2}y^{2}z^{2}=0 (iii) ax^{2}+by^{2}+cz^{2}+2fyz+2gzx+2hxy can be expressed in the form px^{2}+qy^{2}+rz^{2}; and, find p, q, r in terms of a, b, c, f, g, h.


Nalin Pithwa.

Symmetric Functions. Alternating Functions. Algebra for RMO/IITJEE Math

Reference: Higher Algebra by Bernard and Child.

I. Symmetric Functions. 

A function which is unaltered by the interchange of any two of the variables which it contains is said to be symmetric with respect to (wrt) these two variables.

Thus, yz+zx+xy and (x^{2}y+y^{2}z+z^{2}x)(x^{2}z+y^{2}z+z^{2}y) are symmetrical w.r.t. x, y, z.

The interchange of any two letters, x, y, z is called the transposition (xy).

Terms of an expression which are such that one can be changed into the other by one or more transpositions are said to be of the same type. Thus, all the terms of x^{2}y+x^{2}z+y^{2}z+y^{2}x+z^{2}x+z^{2}y are of the same type, and the expression is symmetric with respect to x, y, z.

A symmetric function which is the sum of a number of terms of the same type is often written in an abbreviated form thus: Choose any one of the terms and place the letter \Sigma (sigma) before it. For instance:

x+y+z is represented by \Sigma{x} and xy+yz+zx by \Sigma{xy}.

Again, (x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2xy+2yz+2zx=\Sigma{x^{2}}+2\Sigma{xy}.

It is obvious that

(i) if a term of some particular type occurs in a symmetric function, then all terms of the same type will also occur.

(ii) The sum, difference, product and quotient of two symmetric functions are also symmetric functions.

(PS: there is no need for a grand proof of the above; just apply the definitions of symmetric functions…check with some examples).

Considerations of symmetry greatly facilitate many algebraical processes as illustrated in the following examples:

Example 1: Expand (y+z-x)(z+x-y)(x+y-z).

Solution 1:

This expression is symmetric, homogeneous and of the third degree in x, y, z We may therefore assume that

(y+z-x)(z+x-y)(x+y-z)= a.\Sigma{x^{3}}+b.\Sigma{x^{2}y}+cxyz, where a, b, c are independent of x, y, z. In this assumed identity:

(i) put x=1, y=0, z=0, then -1=a

(ii) put x=1, y=1, z=0, then 0=2a+2b, and so b=1.

(iii) put x=1, y=1, z=1, then 1=3a+6b+c and so c=-2.

Hence, the required product is -x^{2}-y^{2}-z^{2}+y^{2}z+yz^{2}+z^{2}x+zx^{2}+x^{2}y+xy^{2}-2xyz.

Example 2: Expand (a+b+c+d)(ab+ac+ad+bc+bd+cd). Test the result by putting a=b=c=d=1.

Solution 2: the product is the sum of all terms of the product obtained by multiplying any one term of the first expression by any other term of the second expression. Hence, the result will have terms of the type : a^{2}b, abc,

The coefficient of a^{2}b in the product is 1; because this term is obtained as product of a and ab and in no other way.

The coefficient of abc is 3; because this term is obtained in each of the three ways a(bc), b(ca), c(ab).

Hence, the required answer is \Sigma{a^{2}b}+3\Sigma{abc}

Test: The number of terms of the type a^{2}b is 12 and the number of terms of the type abc is 4; hence, if a=b=c=d=1, then \Sigma{a}.\Sigma{ab}=4.6=24 and \Sigma{a^{2}b}+3.\Sigma{abc}=12+3.4=24 so that the test is satisfied.

Example 3: 

Factorize (x+y+z)^{5}-x^{5}-y^{5}-z^{5}.

Solution 3:

Method I: Brute force is really difficult (I did give it a shot …:-))

Method II; Some of you might try the binomial theorem for positive integral index, but to extract the factors is still ..a little bit like brute force method only.

Method III:

Check whether the given expression is symmetric wrt any two variables (namely, x & y; y & z; z & x; ) and whether it is homogeneous and if so, what is the degree. Also from observations of past solved problems, we need to check how many terms of each type are there:

Observations are as follows: the degree of the expression is five only; and the expression is also homogeneous with each term being of degree five; to check for symmetry, let us proceed as follows:

E_{1}=(x+y+z)^{5}-x^{5}-y^{5}-z^{5} and switching x and y gives us E_{2}=(y+x+z)^{5}-y^{5}-x^{5}-z^{5}. Quite clearly, the expression is symmetric w.r.t. x and y; y and z; and, z and x.

To factorize it, we use fundamental theorem of algebra or factor or remainder theorem. Substitute x=-y so that the expression is equal to (z)^{5}-x^{5}-(-x)^{5}-z^{5}=z^{5}-x^{5}+x^{5}-z^{5}=0 so that (x+y) is a factor of the expression. Similarly, the other factors are (y+z) and (z+x). By the fundamental theorem of algebra, we still need a quadratic factor of x, y and z. This factor should be homogeneous also. Hence, let

E=(x+y+z)^{5}-x^{5}-y^{5}-z^{5}=(x+y)(y+z)(z+x){A(x^{2}+y^{2}+z^{2})+B(xy+yz+zx)}, where A and B are pure numeric coefficients independent of x, y and z.

So, put x=1,y=1, z=0, then 2A+B=15 and put x=1, y=1, z=1, then a+b=10 so that A=B=5.

So, E=(x+y+z)^{5}-x^{5}-y^{5}-z^{5}=5(x+y)(y+z)(z+x)(x^{2}+y^{2}+z^{2}+xy+yz+zx).

II Alternating Functions:

If a function E of x, y, z …is transformed into -E by the interchange of any two of the set x, y, z, …, then E is called an alternating function of x, y, z…(Note that just as in the case of symmetric functions, we talk of alternating functions w.r.t. a pair of variables at a time.)

((PS: At this juncture, it behooves you to recall the definitions of even and odd functions, and also to recall the fact that every function can be expressed as a sum of an even function and an odd function. Compare all three now: symmetric, alternating and even/odd functions. ))

Such an alternating function is x^{n}(y-z)+y^{n}(z-x)+z^{n}(x-y); for, the interchange of any two letters, say x and y, transforms it into


Observe that the product and the quotient of two alternating functions are symmetric functions. (here, again, it does not require any grand proof…just pore over the definitions in your head…)

Thus, \frac{x^{3}(y-z)+y^{3}(z-x)+z^{3}(x-y)}{(y-z)(z-x)(x-y)} is symmetric w.r.t. x, y, and z. (PS: please do some scribbling and verify this little observation/fact).

Example 1:

Factorize : x^{3}(y-z)+y^{3}(z-x)+z^{3}(x-y).

Solution 1: 

Let E=x^{3}(y-z)+y^{3}(z-x)+z^{3}(x-y)

We know that E=0 when x=y, y=z and z=y. Hence, the following is a factor of E: (x-y)(y-z)(z-x). As the given expression E is homogeneous of degree 4, it should have one more homogeneous linear factor. The only such factor possible is K(x+y+z). So, now,

E=x^{3}(y-z)+y^{3}(z-x)+z^{3}(x-y)=K(x-y)(y-z)(z-x)(x+y+z). To find K, the numerical coefficient independent of x, y, z, let us equate the coefficient of x^{3}y on each side; then, K=-1. (Alternatively, we could have substituted some numerical values for x, y, z and found K as it is an identity.)

Hence, E=x^{3}(y-z)+y^{3}(z-x)+z^{3}(x-y)=-(x-y)(y-z)(z-x)(x+y+z).

III. Cyclic Expressions:

An algebraic expression is said to be cyclic with respect to the letters a, b, c, d, …, h, k arranged in this order when it remains the same if we replace a by b, b by c, c by d, …., h by k, and k by a.

This “cycle of interchange of letters” is called the cyclic substitution denoted by (abcd\ldots hk).

(PS this reminds you of the the right hand unit vectors i, j, k and their cross products).

Thus, the expression a^{2}b+b^{2}c+c^{2}d+d^{2}a is cyclic with respect to a, b, c, and d (in this order only) because the cyclic substitution (abcd) changes the first term to the second term, the second term to the third term and the fourth term to the first term.

It is clear that:

(i) If a term of some particular type occurs in a cyclic expression, then the term which can be derived from this by the cyclic interchange, must also occur; and, the coefficients of these terms must be equal.

(ii) The sum, difference, product and quotient of two cyclic expressions is also cyclic.

In writing a cyclic expression, it is unnecessary to write the whole expression or all the terms explicitly. Thus, instead of writing the full x^{2}(y-z)+y^{2}(z-x)+z^{2}(x-y) it suffices just to abbreviate it as \Sigma{x^{2}(y-z)}. (Please note that the use of \Sigma here has a different meaning than earlier.)

Sometimes, it is also written in short as x^{2}(y-z)+\ldots+\ldots.

We need to be familiar with the following important basic cyclic identities:

  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^{3}(a-b)=-(b-c)(c-a)(a-b).
  4. bc(b-c)+ca(c-a)+ab(a-b)=-(b-c)(c-a)(a-b).
  5. a(b^{2}-c^{2})+b(c^{2}-a^{2})+c(a^{2}-b^{2})=-(b-c)(c-a)(a-b).
  6. a^{3}(b-c)+b^{3}(c-a)+c^{3}(a-b)=-(b-c)(c-a)(a-b)(a+b+c).
  7. (a+b+c)(ab+bc+ca)=a(b^{2}+c^{2})+b(c^{2}+a^{2})+c(a^{2}+b^{2})+3abc
  8. (b+c)(c+a)(a+b)=a(b^{2}+c^{2})+b(c^{2}+a^{2})+c(a^{2}+b^{2})+2abc
  9. a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca).
  10. (b-c)^{2}+(c-a)^{2}+(a-b)^{2}=2(a^{2}+b^{2}+c^{2}-ab-bc-ca)
  11. (a+b+c)(b+c-a)(c+a-b)(a+b-c)=-a^{4}-b^{4}-c^{4}+2b^{2}c^{2}+2c^{2}a^{2}+2a^{2}b^{2}.

Note that identity 9 can be proved by at least three non-trivial ways 🙂

PS again :-)) It helps to prove the above identities from LHS to RHS and also from RHS to LHS !!

it will be proved later that

i) Any symmetric function of \alpha, \beta, \gamma can be expressed in terms of \Sigma{\alpha}, \Sigma{\alpha\beta} and \alpha\beta\gamma.

ii) Any symmetric function of \alpha, \beta, \gamma, \delta can be expressed in terms of \Sigma{\alpha}, \Sigma{\alpha\beta}, \Sigma{\alpha\beta\gamma}, and \alpha\beta\gamma\delta.

In the above two cases, the notation \Sigma is used in the sense of a symmetric function.

This mode of expression is extremely useful in factorizing symmetric functions, and in proving identities:

Example 1:

Factorize a(1-b^{2})(1-c^{2}) + b(1-c^{2})(1-a^{2}) + c(1-a^{2})(1-b^{2})-4abc.

Solution 1:

PS: Comment: this is not easy. But, go through it and there is ample scope to improve via exercises in the next blog 🙂

Denoting the given expression by E, we have


E=\Sigma{a}-\Sigma{ab^{2}}+abc\Sigma{ab}-4abc, but from identity 7 above, we see that \Sigma{ab^{2}}=\Sigma{a}.\Sigma{ab}-3abc;

Hence, we get E = \Sigma{a}-\Sigma{a}.\Sigma{ab}-abc+abc\Sigma{ab}

So, E=\Sigma{a}.(1-\Sigma{ab})-abc(1-\Sigma{ab})=(1-bc-ca-ab)(a+b+c-abc).

IV. Substitutiions:

We consider processes by which one arrangement (permutation) of a set of elements may be transformed into another:

Taking the permutations cdba, bdac of a, b, c, d, the first is changed into the second by replacing a by c, b by a, c by b and leaving a unaltered. This process is represented by the operator

\left(\begin{array}{cccc}abcd\\cabd \end{array}\right) or \left(\begin{array}{ccc}abc\\cab\end{array}\right)

and, we write \left(\begin{array}{ccc}abc\\cab\end{array}\right)cdba=bdac.

Such a process and also the operator which affects it is called a substitution.

As previously stated, the interchange of two elements a, b  is called the transposition (ab).

Also, a substitution such as \left(\begin{array}{cccc}abcd\\bcda\end{array}\right) in which each letter is replaced by the one immediately following it and the last by the first, is called a cyclic substitution or cycle, and is denoted by (abcd).

If two operators are connected by the sign =, the meaning is that one is equivalent to the other, thus (abcd)=(bcda).

Two or more substitutioins may be applied successively. This is indicated as follows, the order of operations being from right to left.

Let S=(ab), T=(ba), then STabcd=Sacbd=bcad and TSabcd=Tbacd=cabd. Thus, ST=\left(\begin{array}{cccc}abcd\\bcad\end{array}\right) and TS=\left(\begin{array}{cccc}abcd\\cabd\end{array}\right)

This process is called multiplication of substitutions, and the resulting substitution is called the product.

Multiplication of this kind is not necessarily commutative, but if the substitutions have no common letter, it is commutative.

The operation indicated by (ab)(ab), in which (ab) is performed twice, produces no change in the order of the letters, and is called an identical substitution.

Any substituion is cyclic or is the product of two or more cyclic substitutions which have no common element. As an instance, consider the substitution S = \left(\begin{array}{ccccccccc}abcdefghk\\chfbgaedk\end{array}\right)

Here, a is changed to c, c to f, f to a, thus completing the cycle (acf). Also, b is changed to h, h to d, d to b, making the cycle (bhd). Next, c is changed to g, and g to e, giving the cycle (eg). The element k is unchanged, and we write

S=(acf)(bhd)(eg)(k) or S=(acf)(bhd)(eg).

This expression for S in unique, and the order of the factors is indifferent. Moreover, the method applies universally, for in effecting any substitution, we must arrive at a stage when some letter is replaced by the first, thus completing a cycle. The same argument applies to the set of letters not contained in the cycle.

A cyclic substitution of n elements is the product of (n-1) transpositions:


(abc)=(ab)(bc), (abcd)=(abc)(cd)=(ab)(bc)(cd), (abcde)=(abcd)(de)=(ab)(bc)(cd)(de), and so on.

We also have equalities such as : (ae)(ad)(ac)(ab)=(abcde) and (ab)(ac)(ad)(ae)=(edcba).

A substitution which deranges n letters and which is the product of r cycles is equivalent to (n-r) transpositions.

This follows at once from our previous work. Thus, if S = \left(\begin{array}{cccccccc}abcdefgh\\chfbgaed\end{array}\right), then


If we introduce the product (ab)(ab), S is unaltered and the number of transpositions is increased by 2.

Thus, if a given substitutition is equivalent to j transpositions, the number j is not unique. We shall prove that : j=n-r+2s where r is a positive integer or zero.

This is a very important theorem, and to prove it we introduce the notion of “inversions.”

*** Taking the elements a, b, c, d, e choose some arrangement, as abcde, and call it a normal arrangement.

Consider the arrangement bdeac. Here b precedes a, but follows it in the normal arrangement. On this account, we say that the pair ba constitutes an inversion. 

Thus, bdeac contains five inversions, namely, ba, da, dc, ea, ec.

Theorem 1:

If i is the number of inversions which are introduced or removed by a substitution which is equivalent to j transpositions, then i and j are both even or both odd.

Proof of theorem 1:

Consider the effect of a single transposition (fg).

If f, g are consecutive elements, the transposition (fg) does not alter the position of f or g relative to the other elements. It therefore introduces or removes a single inversion due to the interchange of f, g.

If f, g are separated by n elements p, q, r, …, x, then f can be moved to the place occupied by g by n+1 interchanges of consecutive elements, and then g can be moved to the place originally occupied by f by n such interchanges.

Thus, the transposition (fg) can be effected by 2n+1 interchanges of consecutive elements. Therefore, any transposition introduces or removes an odd number of inversions, and the theorem follows. QED.

Again, for a given substitution, i is a fixed number, and therefore whatever value j may have, it must be even or odd, according as i is even or odd. Hence, we get the following:

Theorem 2:

If one arrangement A of a given set of elements is changed into another B by j transpositions, then j is always even or always odd. In other words: the number of transpositions which are equivalent to a given substitution is not unique, but is always even or always odd.

The minimum value of j is n-r.

Thus, substitutions may be divided into two distinct classes. We say that a substitution is even or odd according as it is equivalent to an even or an odd number of transpositions.


To determine the class of a substitution S we may express it as the product of cycles, and count the number of cycles with an even number of elements: then S is even or odd according as this number is even or odd.

Or, we can settle the question by counting the number of inversions, but this generally takes longer.

The tutorial exercises follow this blog.


Nalin Pithwa

Number theory: let’s learn it the Nash way !

Reference: A Beautiful Mind by Sylvia Nasar.

Comment: This is approach is quite similar to what Prof. Joseph Silverman explains in his text, “A Friendly Introduction to Number Theory.”

Peter Sarnak, a brash thirty-five-year-old number theorist whose primary interest is the Riemann Hypothesis, joined the Princeton faculty in the fall of 1990. He had just given a seminar. The tall, thin, white-haired man who had been sitting in the back asked for a copy of Sarnak’s paper after the crowd had dispersed.

Sarnak, who had been a student of Paul Cohen’s at Stanford, knew Nash by reputation as well as by sight, naturally. Having been told many times Nash was completely mad, he wanted to be kind. He promised to send Nash the paper. A few days later, at tea-time, Nash approached him again. He had a few questions, he said, avoiding looking Sarnak in the face. At first, Sarnak just listened politely. But within a few minutes, Sarnak found himself having to concentrate quite hard. Later, as he turned the conversation over in his mind, he felt rather astonished. Nash had spotted a real problem in one of Sarnak’s arguments. What’s more, he also suggested a way around it. “The way he views things is very different from other people,” Sarnak said later. ‘He comes up with instant insights I don’t know I would ever get to. Very, very outstanding insights. Very unusual insights.”

They talked from time to time. After each conversation, Nash would disappear for a few days and then return with a sheaf of computer printouts. Nash was obviously very, very good with the computer. He would think up some miniature problem, usually very ingeniously, and then play with it. If something worked on a small scale, in his head, Sarnak realized, Nash would go to the computer to try to find out if it was “also true the next few hundred thousand times.”

{What really bowled Sarnak over, though, was that Nash seemed perfectly rational, a far cry from the supposedly demented man he had heard other mathematicians describe. Sarnak was more than a little outraged. Here was this giant and he had been all but forgotten by the mathematics profession. And the justification for the neglect was obviously no longer valid, if it had ever been.}


Nalin Pithwa

PS: For RMO and INMO (of Homi Bhabha Science Foundation/TIFR), it helps a lot to use the following: (it can be used with the above mentioned text of Joseph Silverman also): TI nSpire CAS CX graphing calculator.

A fifth primer: plane geometry tutorial for preRMO and RMO: core stuff

  1. Show that three straight lines which join the middle points of the sides of a triangle, divide it into four triangles which are identically equal.
  2. Any straight line drawn from the vertex of a triangle to the base is bisected by the straight line which joins the middle points of the other sides of the triangle.
  3. ABCD is a parallelogram, and X, Y are the middle points of the opposite sides AD, BC: prove that BX and DY trisect the diagonal AC.
  4. If the middle points of adjacent sides of any quadrilateral are joined, the figure thus formed is a parallelogram. Prove this.
  5. Show that the straight lines which join the middle points of opposite sides of a quadrilateral bisect one another.
  6. From two points A and B, and from O the mid-point between them, perpendiculars AP, and BQ, OX are drawn to a straight line CD. If AP, BQ measure respectively 4.2 cm, and 5.8 cm, deduce the length of OX. Prove that OX is one half the sum of AP and BQ. or \frac{1}{2}(AP-BQ) or \frac{1}{2}(BQ-AP) according as A and B are on the same side or on opposite sides of CD.
  7. When three parallels cut off equal intercepts from two transversals, prove that of three parallel lengths between the two transversals the middle one is the Arithmetic Mean of the other two.
  8. The parallel sides of a trapezium are a cm and b cm respectively. Prove that the line joining the middle points of the oblique sides is parallel to the parallel sides, and that its length is \frac{1}{2}(a+b) cm.
  9. OX and OY are two straight lines, and along OX five points 1,2,3,4,5 are marked at equal distances. Through these points parallels are drawn in any direction to meet OY. Measure the lengths of these parallels : take their average and compare it with the lengths of the third parallel. Prove geometrically that the third parallel is the mean of all five.
  10. From the angular points of a parallelogram perpendiculars are drawn to any straight line which is outside the parallelogram : prove that the sum of the perpendiculars drawn from one pair of opposite angles is equal to the sum of those drawn from the other pair.  (Draw the diagonals,and from their point of intersection suppose a perpendicular drawn to the given straight line.)
  11. The sum of the perpendiculars drawn from any point in the base of an isosceles triangle to the equal to the equal sides is equal to the perpendicular drawn from either extremity of the base to the opposite side. It follows that the sum of the distances of any point in the base of an isosceles triangle from the equal sides is constant, that is, the same whatever point in the base is taken).
  12. The sum of the perpendiculars drawn from any point within the an equilateral triangle to the three sides is equal to the perpendicular drawn from any one of the angular points to the opposite side, and is therefore, constant. Prove this.
  13. Equal and parallel lines have equal projections on any other straight line. Prove this.

More later,


Nalin Pithwa.

A fourth primer: plane geometry question set including core theorems, preRMO and RMO

Hard core definitions of various special quadrilaterals:

  1. A quadrilateral is a plane figure bounded by four straight lines.
  2. A parallelogram is a quadrilateral whose opposite sides are parallel.
  3. A rectangle is a parallelogram which has one of its angles a right angle.
  4. A square is a rectangle which has two adjacent sides equal.
  5. A rhombus is a quadrilateral which has all its sides equal, but its angles are not right angles.
  6. A trapezium if a quadrilateral which has one pair of parallel sides.

Problem 1:

The straight lines which join the extremities of two equal and parallel straight lines towards the same parts are themselves equal and parallel. Prove this.

Problem 2:

The opposite sides and angles of a parallelogram are equal to one another, and each diagonal bisects the parallelogram.

Problem 3:

Corollary 1 of problem 2 above: If one angle of a parallelogram is a right angle, all its angles are right angles.

Corollary 2 of problem 2 above: All the sides of a square are equal; and all its angles are right angles.

Corollary 3 of problem 3 above: The diagonals of a parallelogram bisect one another.

Problem 4:

If the opposite sides of a quadrilateral are equal, then the figure is a parallelogram.

Problem 5:

If the opposite angles of a quadrilateral are equal, then the figure is a parallelogram.

Problem 6:

If the diagonals of a quadrilateral bisect each other, then the figure is a parallelogram.

Problem 7:

The diagonals of a rhombus bisect each other at right angles.

Problem 8:

If the diagonals of a parallelogram are equal, all its angles are right angles.

Problem 9:

In a parallelogram which is not rectangular, the diagonals are not equal.

Problem 10:

Any straight line drawn through the middle point of a diagonal of a parallelogram and terminated by a pair of opposite sides is bisected at that point.

Problem 11:

In a parallelogram, the perpendiculars drawn from one pair of opposite angles to the diagonal which joins the other pair are equal. Prove this.

Problem 12:

If ABCD is a parallelogram, and X, Y respectively the middle points of the sides AD, BC, show that the figure AYCX is a parallelogram.

Problem 13:

ABC and DEF are two triangles such that AB, BC are respectively equal to and parallel to DE, EF; show that AC is equal and parallel to DF.

Problem 14:

ABCD is a quadrilateral in which AB is parallel to DC, and AD equal but not parallel to BC; show that (i) angle A + angle C = 180 degrees = angle B + angle D; (ii) diagonal AC = diagonal BD (iii) the quadrilateral is symmetrical about the straight line joining the middle points of AB and DC.

Problem 15:

AP, BQ are straight rods of equal length, turning at equal rates (both clockwise) about two fixed pivots A and B respectively. If the rods start parallel but pointing in opposite senses, prove that (i) they will always be parallel (ii) the line joining PQ will always pass through a certain fixed point.

Problem 16:

A and B are two fixed points, and two straight lines AP, BQ, unlimited towards P and Q, are pivoted at A and B. AP, starting from the direction AB, turns about A clockwise at the uniform rate of 7.5 degrees a second; and BQ, starting simultaneously from the direction BA, turns about B counter-clockwise at the rate of 3.75 degrees a second. (i) How many seconds will elapse before AP and BQ are parallel? (ii) Find graphically and by calculation the angle between AP and BQ twelve seconds from the start. (iii) At what rate does this angle decrease?

Problem 17 (Intercept theorem or Basic Proportionality Theorem):

If there are three or more parallel straight lines, and the intercepts made by them on any transversal are equal, then the corresponding intercepts on any other transversal are also equal.

Prove the corollary: In a triangle ABC, if a set of lines Pp, Qq, Rr, \ldots, drawn parallel to the base, divide one side AB into equal parts they also divide the other side AC into equal parts.


If from the extremities of a straight line AB perpendiculars AX, BY are drawn parallel to a straight line PQ of indefinite length, then XY is said to be the orthogonal projection of AB on PQ.

Problem 18:

Prove: The straight line drawn through the middle point of a side of a triangle parallel to the base bisects the remaining side.

Problem 19:

The straight line which joins the middle points of two sides of a triangle is equal to half the third side.

More later,

Nalin Pithwa.