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 third primer: plane geometry question set for preRMO and RMO

Problem 1:

Prove: Two right angled triangles which have their hypotenuses equal, and one side of one equal to one side of the other, are equal in all respects. (PS: Please do not use “short-cut or the magic formula — Pythagoras’s theorem; if you do, you are requested to prove Pythagoras’s theorem using plane geometry!) (PS:2: Remember Euclid’s geometry builds up from “scratch”…a small step at a time….axioms, definitions, lemmas, …; you can only use the theorems used and proved so far in “class”).

Problem 2:

If two triangles have two sides of the one equal to two sides of the other, each to each, but the angle included by the two sides of one greater than the angle included by the corresponding sides of the other; then the base of that which has the greater angle is greater than the base of the other.

Problem 3:

Prove converse of the above: If two triangles have two sides of the one equal to two sides of the other, each to each, but the base of one greater than the base of the other, then the angle contained by the sides of that which has the greater base is greater than the angle contained by the corresponding sides of the other.

Problem 4:

(i) The perpendicular is the shortest line that can be drawn to a given straight line from a given point. (ii) Obliques which make equal angles with the perpendicular are equal. (iii) Of two obliques the less is that which makes the smaller angle with the perpendicular. Prove all these.

Problem 5:

If two triangles have two sides of the one equal to two sides of the other, each to each, and have likewise the angles opposite to one pair of equal sides equal, then the angles opposite to the other pair of equal sides are either equal or supplementary, and in the former case, the triangles are equal in all respects.

More later,


Nalin Pithwa

A Second primer: plane geometry for preRMO and RMO: basic questions set

Problem 1:

If from any point in the bisector of an angle a straight line is drawn parallel to either arm of the angle, the triangle thus formed is isosceles.

Problem 2:

From X, a point in the base BC of an isosceles triangle ABC. a straight line is drawn at right angles to the base, cutting AB in Y, and CA produced in Z, show the triangle AYZ is isosceles.

Problem 3:

If the straight line which bisects an exterior angle of a triangle is parallel to the opposite side, show that the triangle is isosceles.

Problem 4:

The straight line drawn from any point in the bisector of an angle parallel to the arms of the angle, and terminated by them, are equal, and the resulting figure is a quadrilateral having all its sides equal.

Problem 5:

AB and CD are two straight lines intersecting at D, and the adjacent angles so formed are bisected; if through any point X in DC a straight line XYZ is drawn parallel to AB and meeting the bisectors Y and Z, show that XY is equal to XZ.

Problem 6:

Two straight rods PA, QB revolve about pivots at P and Q, PA making 12 complete revolutions a minute, and QB making 10. If they start parallel and pointing the same way, how long will it be before they are again parallel (i) pointing opposite ways (ii) pointing the same way?

Problem 7:

Prove that: if two straight lines are perpendicular to two other straight lines, each to each, the acute angle between the first pair is equal to the acute angle between the second pair.

Problem 8:

Show that the only regular figures which may be fitted together so as to form a plane surface are (i) equilateral triangles (ii) squares (iii) regular hexagons.

Problem 9:

If one side of a regular hexagon is produced, show that the exterior angle is equal to the interior angle of an equilateral triangle.

Problem 10:

If a straight line meets two parallel straight lines, and the two interior angles on the same side are bisected, show that the bisectors meet at right angles.

Problem 11:

If the base of any triangle is produced both ways, show that the sum of the two exterior angles minus the vertical angle is equal to two right angles.

Problem 12:

In the triangle ABC, the base angles at B and C are bisected by BO and CO respectively. Show that the angle BOC is 90 degrees plus A/2.

Problem 13:

In the triangle ABC, the sides AB, AC are produced, and the exterior angles are bisected by BO and CO. Show that the angle BOC is 90 degrees minus A/2.

Problem 14:

Prove: the angle contained by the bisectors of two adjacent angles of a quadrilateral is equal to half the sum of the remaining angles.

Problem 15:

A is the vertex of an isosceles triangle ABC, and BA is produced to D, so that AD is equal to BA; if DC is drawn, show that BCD is a right angle. Prove this.

Problem 16:

Prove: The straight line joining the middle point of the hypotenuse at a right angled triangle to the right angle is equal to half the hypotenuse.

Problem 17:

If two triangles have two angles of one equal to two angles of the other, each to each, and any one side of the first equal to the corresponding side of the other, the triangles are equal in all respects. (ASA congruency test):

Problem 18:

Show that the perpendiculars drawn from the extremities of the base of an isosceles triangle to the opposite sides are equal.

Problem 19:

Prove: Any point on the bisector of an angle is equidistant from the arms of the angle.

Problem 20:

Through O the middle point of a straight line AB, any straight line is drawn, and perpendiculars AX and BY are dropped upon it from A and B: show that AX is equal to BY.

Problem 21:

If the bisector of the vertical angle of a triangle is at right angles to the base, the triangle is isosceles.

Problem 22:

If in a triangle the perpendicular from the vertex on the base bisects the base, then the triangle is isosceles. Prove.

Problem 23:

If the bisector of the vertical angle of a triangle also bisects the base, the triangle is isosceles. Prove this.

Problem 24:

The middle point of any straight line which meets two parallel straight lines, and is terminated by them, is equidistant from the parallels. Prove this.

Problem 25:

A straight line drawn between two parallels and terminated by them is bisected; show that any other straight line passing through the middle point and terminated by the parallels is also bisected at that point.

Problem 26:

If through a point equidistant from two parallel straight lines, two straight lines are drawn cutting the parallels, the portions of the latter thus intercepted are equal. Prove this.

Problem 27:

In a quadrilateral ABCD, if AB=CD, and BC=DC: show that the diagonal AC bisects each of the angles which it joins, and that AC is perpendicular to BD.

Problem 28:

A surveyor wishes to ascertain the breadth of a river which he cannot cross. Standing at a point A, near the bank, he notes an object B immediately opposite on the other bank. He lays down a line AC of any length at right angles to AB, fixing a mark at O the middle point of AC. From C he walks along a line perpendicular to AC until he reaches a point D from which O and B are seen in the same direction. He now measures CD: prove that the result gives him the width of the river.

More later,


Nalin Pithwa.

PS: There is no royal road to geometry —- Plato.