Looking for a Tutor Near You?

Post Learning Requirement »
x

Choose Country Code

x

Direction

x

Ask a Question

x

Hire a Tutor
Dubai

Notes On Topics Of Algebra

Loading...

Published in: Mathematics
969 Views

Attached files contain the basic concept of Algebra topics.

Himanshu / Dubai

15 years of teaching experience

Qualification: B.Tech. IIT Delhi,

Teaches: SAT, ACT, Maths, IIT JEE, Chemistry, Computer Science, English, French, Physics, Science, Computer, Mathematics, English Language

Contact this Tutor
  1. Basic Principles of Counting FUNDAMENTAL PRINCIPAL OF COUNTING Rule of Product If one experiment has n possible outcomes and another experiment has m possible outcomes, then there are m x n possible outcomes when both of these experiments are performed. In other words if a job has n parts and the job will be completed only when each part is completed and the first part can be completed in al ways, the second part can be completed in a2 ways and so on . . then nth part can be completed in an ways, then the total number of ways of doing the job is ala2a3 . an. This is known as the rule of product. Illustration: A college offers 7 courses in the morning and 5 in the evening. Find the possible number of choices with the student who wants to study one course in the morning and one in the evening. Solution: The student has seven choices from the morning courses out of which he can select one course in 7 ways. For the evening course, he has 5 choices out of which he can select one in 5 ways. Hence the total number of ways in which he can make the choice of one course in the morning and one in the evening = 7 x 5 = 35. Illustration: A man has five friends. In how many ways can be invite one or more of them to a tea pa rty. Solution: Methods I Case I: Case 11: Case 111: Case IV: The man can invite one friend in 5C1 ways The man can invite two friend in 5C2 ways The man can invite three friend in 5C3 ways The man can invite four friend in 5C4 ways
  2. Case V: The man can invite five friend in 5C5 ways All these five cases are mutually exclusive i.e. the man can either invite one friend or two friends or . but can not simultaneously invite one friend and two friends. Remember: Whenever there is 'OR', we add. Therefore Total number of ways in which the man can invite his friends = 5C1 + 5C2 + 5C3 + 5C4 + 5C5 = 5 + 10 + 10 + 5 + 1 = 31 Method 11. This problem can also be solved as follows: Each of his friends may be dealt in two ways - either invited or not invited. Therefore The total number of ways of inviting all friends is 32 ways. But this includes the case, in which all the friends are not invited. Hence, number of ways in which the man invites one or more friends is 25 - 1 = 31 ways. the total Note: In the above illustration, we have found the total number of combinations of n dissimilar things taking any number of them at a time, which is 2n - 1. Rule of Sum If one experiment has n possible outcomes and another has m possible outcomes, then there are (m + n) possible outcomes when exactly one of these experiments is performed. In other words if a job can be done by n methods and by using the first method can be done in al ways or by second method in a2 ways and so on . by the nth method in an ways, then the number of ways to get the job done is (al + a2 + Illustration: How many straight lines can be formed from six points, no three of which are collinear? Solution: To form a straight line, we need to select two points out of six points. This can be done in ways = 6.5/2.1 = 15 ways. Illustration:
  3. In how many ways can a committee of five be formed from amongst four boys and six girls so as to include exactly two girls? Solution: We have to select two girls from six girls and three boys from four boys. Number of ways of selecting girls = 6C2 = 6.5/2.1 = 15 Number of ways of selecting boys = 4C3 = 4 Number of ways of forming the committee = 15 x 4 = 60. Note: Illustration: Here we have used multiplication rule. A college offers 7 courses in the morning and 5 in the evening. Find the number of ways a student can select exactly one course, either in the morning or in the evening. Solution: The student has seven choices from the morning courses out of which he can select one course in 7 ways. For the evening course, he has 5 choices out of which he can select one in 5 ways. Hence he has total number of 7 + 5 = 12 choices. Illustration: How many (i) 5 - digit, (ii) 3 repetition of digits. Solution: - digit numbers can be formed by using 1, 2, 3, 4, 5 without (i) Making a 5-digit number amounts to filling 5 places. Places: Number of Choices: 5 4 3 2 1 The first place can be filled in 5 ways using any of the given digits. The second place can be filled in 4 ways using any of the remaining 4 digits. Similarly, we can fill the 3rd 4th and 5th place.
  4. No. of ways to fill all the five places. - 120 = > 120 5-digit numbers can be formed. Making a 3 - digit number amounts to filling 3 places. Places: Number of choices: 5 4 Number of ways to fill all the three places — Hence the total possible 3 - digit numbers 3 = 60. - 60. Illustration: How many 4-letter words can be formed using a, b, c, d, e (i) Without repetition Solution: (ii) with repetition (i) The number of words that can be formed is equal to the number of ways to fill the three places. Places: Number of Choices: 5 4 3 2 = > 5 x 4 x 3 x = 120 words can be formed when repetition is not allowed. (ii) The number of words that can be formed is equal to the number of was to fill the three places. Places: Number of Choices: 5 5 5 5 First place can be filled in 5 ways. If repetition is allowed, all the remaining places can be filled in 5 ways each. = > 5 x 5 x 5 x 5 = 625 words can be formed when repetition is allowed. Exercise
  5. 1. 2. Ans. 1 -1 1 Plane, 2 trains and 3 buses ply between Delhi and Agra. (a) In how many ways can you to Agra from Delhi. (b) In how many ways can you go and come back if you go by train, Total number of combinations of n dissimilar things, taken at least one at a time - 12 2.
  6. Permutations (Arrangements of Objects): The number of permutations of n objects, taken r at a time, is the total number of arrangements of n objects, in groups of r where the order of the arrangement is important. (i) Without repetition (a) Arranging n objects, taking r at a time in every arrangement, is equivalent to filling r places from n things. r-Places Number of Choices: 1 n 2 3 4 Number of ways of arranging n-2 n-3 n-l = Number of ways of filling r places =npr (b) (a) n(n - l)(n - 2) n!/((n-r)!) Number of arrangements of n different objects taken all at a time = npn With repetition Number of permutations (arrangements) of n different objects, taken r at a time, when each object may occur once, twice, thrice . . up to r times in any arrangements = Number of ways of filling r places, each out of n objects. r-Places: Number of Choices : 1 n 2 n 3 n 4 n n Number of ways to arrange = Number of ways to fill r places Number of arrangements that can be formed using n objects out of which p are identical (and of one kind), q are identical (and of one kind) and rest are different n!/p!q!r!. Illustration: How many 7 - letter words can be formed using the letters of the words: (a) BELFAST, ( ) ALA BA M A
  7. Solution: (a) (b) BELFAST has all different letters. Hence, the number of words = 5040. ALABAMA has 4 A's but the rest are all different. Hence the number of words formed is 7!/4! 210. Illustration: (b) (d) How many anagrams can be made by using the letters of the word HINDUSTAN? How many of these anagrams begin and end with a vowel. In how many of these anagrams all the vowels come together. In how many of these anagrams none of the vowels come together. In how many of these anagrams do the vowels and the consonants occupy the same relative positions as in HINDUSTAN? Solution: (a) The total number of anagrams = Arrangements of nine letters taken all at a time = 9!/2! = 181440. (b) We have 3 vowels and 6 consonants, in which 2 consonants are alike. The first place can be filled in 3 ways and the last in 2 ways. The rest of the places can be filled in 7!/2! ways. Hence the total number anagrams = 3 x 2 x 7!/2! = 15210. (c) Assume the vowels (IUA) as a single letter. The letters (IUA) H, D, S, T, N, N can arranged in 7!/2! ways. Also IUA can be arranged, among themselves, in 3! = 6 ways. Hence the total number of anagram = 7!/2! x 6 = 15120. (d) Let us divide the task in two parts. In the first, we arrange the 6 consonants as shown in 6!/2! ways. x C x C x C x C x C x C x (C stands for consonants and x stand for blank spaces between them) 3 vowels can be arranged in 7 places (between the consonants) in 7P3 In this case the vowels are arranged among themselves in 3! consonants are arranged among themselves in 6!/2! ways. = 7!/2! =210. — 6 ways. Also the
  8. Hence the total number of anagrams Illustration: = 6!/2! x 2160. How many 3 digit numbers can be formed using the digits 0, 1, 2, 3, 4, 5 where digits may not be repeated, (b) digits may be repeated Solution: (a) Let the 3 digit number by XYZ. Position (X) can be filled by 1, 2, 3, 4, 5 but not 0, so it can be filled in 5 ways. Position (Y) can be filled in 5 ways again. Since 0 can be placed in this position. Position (Z) can be filled in 4 ways. Hence, by the fundamental principal of counting, total number of ways is 5 x 5 x 4 = 100 ways. (b) Let the 3 digit number be XYZ. Position (X) can be filled in 5 ways. Position (Y) can be filled in 6 ways Position (Z) can be filled in 6 ways. Hence, by the fundamental principle of counting, total number of ways is 5 x 6 x 6 = 180. Illustration: Find the number of ways in which 6 letters can be posted in 10 letterboxes. Solution: For every letter, Hence the total we have 10 choices (i.e. 10 letterboxes). = 106 number of ways
  9. Circular Permutations The arrangements we have considered so far are linear. There are also arrangements in closed loops, called circular arrangements. Consider four persons A, B, C and D, who are to be arranged along a circle. It's one circular arrangement is as shown in adjoining figure. c Shifting A, B, C, D one position in anticlockwise direction we will get arrangements as follows. c Arrangements as shown in figure position of none of the four persons arrangements the four arrangements are. ABC D; D (1) (11) (111) CD A B; and is c (IV) are not different as relative changed. But in case of linear B C DA Thus, it is clear that corresponding to a single circular arrangement of four different things there will be 4 different linear arrangements. Let the number of different things be n and the number of their circular permutations be x. Now for one circular permutation, number of linear arrangements is n For x circular arrangements number of linear arrangements — nx.
  10. But number of linear arrangements of n different things From (1) and (2) we get Suppose n persons (al, a2, a3, . Nx = n! = > x = n!/n , an) are to be seated around a circular table. There are n! ways in which they can be seated in a row. On the other hand, all the linear arrangements al, a2, a3, . an, al, a2, • an-I, an, al, a2, • , an , an-I al will lead to the same arrangements for a circular table. Hence one circular arrangement corresponds to n unique row (linear) arrangements. Hence the total number of circular arrangements of n persons IS In other words the permutation in a row has a beginning and an end, but there is nothing like beginning or end in circular permutation. Hus, in circular permutation, we consider one object is fixed and the remaining objects are arranged in (n - 1)! ways (as in the case of arrangement in a row). Distinction between clockwise and Anti-clockwise Arrangements Consider the following circular arrangements: 1 11 In figure I the order is clockwise whereas in figure Il, the other is anti-clock wise. These are two different arrangements. When distinction is made between the clockwise and the anti- clockwise arrangements of n different objects around a circle, then the number of arrangements
  11. But if no distinction is made between the clockwise and anti-clockwise arrangements of n different objects around a circle, then the number of arrangements is (n - 1)! . As an example consider the arrangements of beads (all different) on a necklace as shown in figure A and B. FLIP TO RIGHT Look at (A) having 3 beads as shown. Flip (A) over on its right. We get (B) at once. However, (A) and (B) are really the outcomes of one arrangement but are counted as 2 different arrangements in our calculation. To nullify this redundancy, the actual number of different arrangements is Note: (i) When the positions are numbered, circular arrangements is treated linear arrangement. as a In linear arrangements it does not make difference whether the positions are numbered or not. Illustration: 20 persons we invited to a party. In how many ways can they be seated in a round table such that two particular persons sit on either side of the host? Solution: After fixing the places of three persons (1 host + 2 persons) and treating them as 1 unit we can arrange the total (20 - 2 + 1) — 19 units in 18! ways. Again these particular persons can sit on either side of the host in 2 ways. Hence the total number of ways is 18! x 2. Illustration: In how many ways 10 boys and 5 girls can sit around a circular table, so that no two girls sit together. Solution:
  12. stands for boys) 10 boys can be seated in a circle in 9! ways. There are 10 spaces between the boys, which can be occupied by 5 girls in 10P5 ways. Hence total number of ways = 9! 10P5 Number of circular permutations of n different things taken r at a time r (if clockwise and anticlockwise orders are taken as different) r (if clockwise and anticlockwise orders are not taken to be different) Illustration: In how many ways can 20 persons be seated round a table if there are 9 chairs? Solution: In case of circular table the clockwise and anticlockwise arrangements are different. Hence the total number of ways — Illustration: 9 How many necklaces of 10 beads each can be made from 20 beads of different colours? Solution: In case of necklace there is no distinction between the clockwise and arrangements. Then the required number of circular permutations anticlockwise
  13. Combinations Meaning of combination is selection of objects. Selection of objects without repetition: The number of selections (combinations or groups) that can be formed form n different objects taken r(0 S r S n) at a time is ncr = n!/(r!(n-r))!. Explanation: Let the total number of selection (or groups) — x. Each group contains r objects, which can be arranged in r! ways. Hence number of arrangements of r objects = x But the number of arrangements = = = > x = n!/(r!(n-r)!) Selection of objects with repetition ncr. The number of combination of n distinct objects taken r at a time when each may occur — n+r-1C once, twice, thrice, upto r times, in any combination — Explanation: Let eh n objects al, a2, a3, . al occur Xl times, a2 occur times, a3 occur times, an occur xn times, such that Xl + + n}. . an. In a particular group of r objects, let Now the total number of selections of r objects, out of n number of non-negative integral solution of equation (1)
  14. Details of finding the number of integral solutions of equation (1) are given on Note: page 12 (Multinomial theorem). Illustration: Let 15 toys be distributed among 3 children subject to the condition that any child can take any number of toys. Find the required number of ways to do this if toys are distinct, Solution: if toys are identical (i) Toys are distinct Here we have 3 children and we want the 15 toys to go to the 3 children with repletion. In other words it is same as selecting and arranging children 15 times out of 3 children with the condition that any children can be selected any no. of time which can be done in 315 ways (n = 3, r = 15). (ii) Toys are identical Here we only have to select children 15 times out of 3 children with the condition that any children can be selected any number of times which can be done in 3+5-1C15 = 17C2 way (n = 3, r = 5).
  15. Permutations vs Combinations DIFFERENCE BETWEEN PERMUTATION AND COMBINATION: In combination we are concerned only with the number of things n each selection whereas in permutation we also consider the order of things which makes each arrangement different The symbol npr, denotes the number of permutations of n different things taken r at a time, whereas ncr denotes the number of combinations of n different things taken r at a time. Enquiry: How can we find the number of ways of selection n things taken r at a time, when their order also matters? This can be done by finding the number of permutations of n different things taken r at a time (n > r) or to find the value of npr. This is same as finding the number of ways in which we can fill up r places when we have n different things at our disposal. The first place may be filled up in n-ways, because we may take any one of the n things. When it has been filled up in any one of these ways, the second place can then be filled up in (n - 1) ways, and since each way of filling up the first place can be associated with each way of filling up the second, the number of ways in which the first two places can be filled up is given by the product n(n - 1) ways, And when the first two places have been filled up in any way, the third place can be filled up in (n - 2) ways. And reasoning as before, the number of ways in which three places can be filled up is n(n - Proceeding thus, and noticing that a new factor is introduced with each new place filled up, and that at any stage the number of factors is the same as the number of places filled up, we shall have the number of ways in which r places can be filled up, we shall have the number of ways in which r places can be filled up equal to n(n - l)(n - 2) . . to r factor, and the rth factor is n - (r - 1), or n 1. Therefore the number of permutations of n things taken r at a time is npr — n(n -
  16. Restricted Selection and Arrangement (a) The number of ways in which r objects can be selected form n different objects if k particular objects are — n -kcr-k. always included — n-kc never included = (b) The number of arrangement of n distinct objects taken r at a time so that k particular objects are n-kCr_k. r! , always included = — n-kCr.r!. never included — Illustration: A delegation of four students is to be selected form a total of 12 students. ways can the delegation be selected if In how many (b) (c) (d) all the students are equally willing. two particular students have to be included in the delegation. two particular students do not wish to be together in the delegation. two particular students wish to be included together only, two particular students refuse to be together and two other particular student wish to be together only in the delegation. Solution: (a) Formation of delegation means selection of 4 out of 12. Hence the number of ways — 12C4 = 495. (b) Two particular students are already selected. Hence we need to select 2 out of the remaining 10. Hence the number of ways = IOC2 = 45. The number of ways in which both are selected = 45. Hence the number of ways in which the two are not included together = 495 - 45 = 450. (d) There are two possible cases Either both are selected. In this case the number of ways in which this selection can be made = 45.
  17. or (ii) both are selected. In this case all the four students are selected from the rest of ten students. This can be done in IOC4 = 210 ways. Hence the total number of ways of selection = 45 + 210 = 255. (e) We assume that students A and B wish to be selected together and students C and D do not wish to be together. Now there are following 6 cases. Ck, then r — kor n-r — k (D) selected selected selected selected not selected For (i) the number of ways selection For (ii) the number of ways selection For (iii) the number of ways selection For (iv) the number of ways selection — For (v) the number of ways selection — For (vi) the number of ways selection (D) not selected (C) not selected (C, D) not selected (A, B, D) not selected (A, B, C) not selected = 8C1 = 8 = 8C1 = 8C2 _ 8C3 _ 8C3 = 8C4 = 28 = 56 = 56 = 70 Hence, total number of ways = 8+ 8 + 28 + 56 + 56 + 70 = Some results related to "Cr 226. (iii) If = n — n+1C n Cr + n Cr-l ncr = n/r n-lCr-l
  18. nc (a) (b) If n is even ncr is greatest for r = n/ 2 If n is odd, is greatest for r Restricted Selection and Arrangement Illustration: (a) How many diagonals are there in as n-sided polygon (n > 3? (b) How man y triangles can be formed by joining the vertices of an n-sided polygon? How many these triangles have exactly one side common with that of the polygon? exactly two sides common with that of the polygon? no side common with that of the polygon? (iii) Solution: (b) Number of lines formed by joining the vertices of a polygon = number of selection of 2 points each selected from the given n points = nC2 = . Out of nC2 lines, n are the sides of the polygon. Hence the number of diagonals = nC2-n = ((n(n-1))/2) n Number of triangles formed by joining the vertices of the polygon — number of selection of 3 points from n points.
  19. nC3 = . Let the vertices of the polygon be marked as Al, A2, A3, . (i) Select two consecutive vertices Al, A2 of the polygon. For the required triangles we can select the third vertex from the . An-I. This can be done in n-4C1 ways. Also two consecutive points (end points of a side of polygon) can beselected in n ways. For the required triangle, we have to select three consecutive vertices of the polygon. i.e. (Al A2 A3), (A2 A3 A4) (A3 A4A5) ... (An Al A2). This can be done in n ways. (iii) Triangle having no side common + triangle having exactly one side common + triangle having exactly two sides common (with those of the polygon) = Total number triangles formed = > Triangle having no side common with those of the polygon. = nC3 - n(n - 4) -n = n/6 [n2 + +2- 6n + 24 - 6] = [n2 + 9 + 20] = All possible selections (i) Selection from distinct objects : The number of selections from n different objects, taking at least one = nC1 + nC2 +nC3 + ... + nCn = 2n - 1. In other words, for every object we have two choices, either select or reject in a particular group. Total number of choice (all possible selections) = 2.2.2. ...n times = 2n. But this also includes the case when none is selected. And the number of such cases is 2n - 1. Selection from identical objects: The number of selections of r objects out of n identical objects is 1. Total number of selections of zero or more objects from n identical objects is n+l. (b) (c) The total number of selections of at least one out of al + a2 + a3 + . . + an objects, where al are alike (of one kind), a2 are alike (of second kind) and so on . . an are alike (of nth kind), is (iii) Selection when both identical and distinct objects are present: The number of selections, taking at least one out of al + a2 + a3 + ... an + k objects, where al are alike (of one kind), a2 are alike (of second kind) and so on an are alike (of nth kind), and k are distinct Illustration: In how many ways can a person having 3 coins of 25 paise, 4 coins of 50 paise and 2 coins of 1 rupee give none or some
  20. coins to a beggar? Restricted Selection and Arrangement Solution: The person has 3 coins of 25 paise, 4 coins of 50 paise and 2 coins of 1 rupee. The number of ways in which he can give none or some coins to a beggar is (3 + 1)(4 + 1)(2 + 1) (iv) (a) Total number of divisor's of a given natural number 0.1 — = 60 ways. pn , To find the number of factors of a given natural number greater than 1 we can write n as n = where PI, p2, . be of the form d = pn are distinct prime numbers and al, a2 . . an are non-negative integers. Now any divisor of n will P2 here number of factors will be equal to numbers of ways in which we can choose ßls' which can be done in (al + 1)(a2 + l)......(an+ 1) ways. Sum of all the divisors of n is given by PI-I —1 P2—1 —1 Pz—l (b) The exponent of a prime number pl in n ct2+1 —1 Pn—l pnan is given by
  21. Illustration: How many positive factors are there of the number 360 and find the sum of all these factors. Solution: Let n = 360 = 23.32.5 = > No. of factors of 360 = (3 + + + 1) = 24. Sum of all the factors =((24-1)/1) . ((32-1)/2) . ((52-1)/4) = 15. 13. 6 = 1170.
  22. Division and Distribution of Objects (With fixed number of objects in each group) Into groups of unequal size (different number of objects in each group) (a) Number of ways, in which n distinct objects can be divided into r unequal groups containing al, a2, a3, . things (al * aj) n-a1-a2c n-a ICa2. ncal. ar. Here a 1 + a 2 + a 3 + n. (b) Number of ways in which n distinct objects can be distributed among r persons such that some person get al objects, another person get a2 objects . and similarly someone gets ar objects Explanation: Let us divide the task into two parts. In the first part, we divide the objects into groups. In the second part, these r groups can be assigned to r persons in r! ways. Into groups of equal size (each group containing same number of objects) (a) Number of ways in which m x n distinct objects can be divided equally into n groups (unmarked) = (mn)!/(m!)n n! . (b) Number of ways in which m x n different object can be distributed equally among n persons (or numbered groups) — (number of ways of dividing) x (number of groups)! (mn)!/(m!)n
  23. Derangements and Multinomial Theorem DERANGEMENTS Any change in the existing order of things is called a derangement. If 'n' things are arranged in a row, the number of ways in which they can, that none of them occupies its original place is be deranged so 1 1 and it is denoted by D(n). A question on derangement can be of the following kind: Illustration: 1 Supposing 4 letters are placed in 4 different envelopes. In how many ways can be they be taken out from their original envelopes and distributed among the 4 different envelopes so that no letter remains in its original envelope? Solution: Using the formula for the number of derangements that are possible out of 4 letters in 4 envelopes, we get the number of ways as . -1 + 1/2! - 1/3! + 1/4!) = 24(1 -1 + 1/2 - 1/6 + 1/24) = 9. MULTINOMIAL THEOREM Let ......, Xm be integers. Then number of solutions to the equation subject to the conditions al < Xl < bl, a2 < < b2 is equal to the coefficient of xn in , am < Xm < bm (X am + +xbm) ...(3) This is because the number of ways in which sum of m integers in (1) subject to given conditions (2) equals n is the same as the number of times xn comes in (3). Using this we get the number of non negative integral solutions of (1) is given byn+m-1Cm-1 and number of positive integral solutions of (1) is given by n-1Cm-1.
  24. Illustration: In how many different ways three persons A, respectively can donate Rs.10 collectively. Solution: B, C having 6, 7 and 8 one rupee coins The number of ways in which they can denote Rs.10 is the same as the number of solutions of the equation = 10 Xl + X2 + X3 subject to conditions 0 S Xl S 6, 0 S S 7, 0 S < 8 Hence the required number of ways — coefficient of in (1 = coefficient of x10 in (I-x7)(1-x8)(1-x9)(1-x)-3 = coefficient of x10 in + 5C3x3+...+12C10x10) — 12C2 = 66-10-6-3- (ignoring powers higher than 10) _ 3C1 _ 47.

Need a Tutor or Coaching Class?

Post an enquiry and get instant responses from qualified and experienced tutors.

Post Requirement

Related Study Notes

Upload Study Notes

If you have your own Study Notes which you think can benefit others, please upload on MyPrivateTutor. For each approved Study Note you will get 100 Credit Points and 100 Activity Score which will increase your profile visibility.

Upload Now
Home > Study Notes > Notes On Topics Of Algebra