What is the use of Mathematics

Posted by

0

Mathematics Hothouse

Well, you might be asking this question in high school. You might have found that Math is a lot of formulae and manipulations similar to black magic in Algebra and wild imaginations in Geometry — I mean the proofs. So Math means prove this and that. Right?

I agree to some extent. Initially, it is sort of drab or *mechanical*. But, there is a good analogy. Think how you learnt writing the English alphabet — keep on drawing a big A, retracing it 10 times daily and perhaps, your Mom would have whacked you if you did not want to practise it. But, after you know English, the whole world of opportunities opens up for you; your vistas have widened. Exactly same is the case with Mathematics of high school and junior college. But, of course, there are applications of high school math which you learn later when you pursue…

View original post 565 more words

Probability Theory Primer: part I

Posted by

0

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?

Regards,

Nalin Pithwa.

 

 

 

 

 

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

Posted by

0

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.

Regards,

Nalin Pithwa.

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

Posted by

0

  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)

Regards,

Nalin Pithwa.

 

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

Posted by

0

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}

4.(\alpha\beta+\alpha\gamma+\alpha\delta+\beta\gamma+\beta\delta+\gamma\delta)^{2}=\Sigma{\alpha^{2}\beta^{2}}+2\Sigma{\alpha^{2}\beta\gamma}+6\alpha\beta\gamma\delta

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

Regards,

Nalin Pithwa.

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

Posted by

0

Exercises:

  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.

Cheers,

Nalin Pithwa.