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Branch of Mathematics |
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Ganithguru
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ALGEBRA Series(special)
GEOMETRY Theoretical Geometry Constructions Applied geometry
TRIGONOMETRY Angles and measures Trigonometric Ratio Algebra of T-functions Properties of triangles Applied Trigonometry
CALCULUS Examples of different kinds of functions Graph of functions Limit of a function Continuity of a function Differential Calculus Integral Calculus Applied Calculus
Sample lesson and problems for Competitive exams
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Permutation
, Combination Permutations: Factorial:
5! = 1´2´3´4´5=120 Zero Factorial: We will require zero factorial in the latter sections of this chapter. It does not make any sense to define it as the product of the integers from 1 to zero. So, we define 0! = 1. Fundamental Principles of Counting: Fundamental Principle of Multiplication: If there are two jobs such that one of them can be completed in m ways, and when it has been completed in any one of these m ways, second job can be completed in n ways: then the two jobs in succession can be completed in m x n ways. Fundamental Principle of Addition: If there are two jobs such that they can be performed independently in m and n ways respectively, then either of the two jobs can be performed in (m + n) ways. Concept of Permutations: Definition: The number of arrangements that can be made out of n things taking r at a time is called the number of permutations of n things taken r at a time. Notation: If n and r are positive
integers such that 1 £ r £ n, then the number of all permutations of n distinct things, taken r at a time is denoted by the symbol P(n,r) or nPr. We use the symbol nPr throughout our discussion. Thus nPr = total number of permutations of n distinct things taken r at a time. Note: In permutations the order of arrangement is taken into account; when the order is changed, a different permutation is obtained. Theorem : Let r and n be positive
integers such that 1 £ r £ n. Then the number of all
permutations of n distinct things
taken r at a time is given by n (n -1) (n -2) i.e. nPr = n(n-1)(n-2) …(n-r-1) Permutations of objects not all distinct: The number of mutually
distinguishable permutations of n
things, taken all at a time, of which p
are alike of one kind, q alike of
second such that p+q = n, is Permutations when objects can repeat: The number of permutations of n different things, taken r at a time, when each may be repeated nay number of times in each arrangement, is nr. Circular Permutations: We have seen that the number of permutations of n different things taken all together is n!, where each permutation is a different arrangement of then things in a row, or a straight line. These permutations are called linear premeditations or simply permutations. A circular permutation is one in which the things are arranged along a circle. It is also called closed permutation. Theorem : The number of circular
permutations of n distinct objects
is (n-1)! Note: In the above theorem anti-clockwise and clockwise order of arrangements are considered as distinct permutations. Difference between clock wise and anti-clockwise arrangements: Consider the following circular permutations:
We observe that in both, the
order of the circular arrangement is a1, a2,
The word combinations mean selection. Suppose we are asked to make a selection of any two things from three things a, b and c, the different selections are ab, bc, ac. Here there is no reference to the order in which they are selected. i.e., ab and ba denote the same selection. These selections are called combinations. Definition: A selection of any r things out of n things is called a combination of n things r at a time. Notation: The number of all combinations of n objects, taken r at a time is generally denoted by nCr
or C(n,r) or Thus nCr ={Number of ways of selecting r objects from n objects. Difference between Permutation and Combination: 1.In a combination only selection is made whereas in a permutation not only a selection is made but also an arrangement in a definite order is considered. i.e. in a combination, the ordering of the selected objects is immaterial whereas in a permutation, the ordering is essential. 2.Usually the number of permutation exceeds the number of combinations. 3.Each combination corresponds to many permutations. Combinations of n different
things taken r at a time: Theorem 3.5: The number of all combinations of n distinct objects, taken r
at a time is given by nCr = Problems about Counting 1. How many ways can you arrange 7 different
books, so that a specific book is on the third place? Remove the third book. Now you can arbitrary arrange the 6 books in 6! ways. 2. In how many ways can you take 3
marbles out of a box with 15 different marbles? Taking a subset of three elements out a set of 15 elements can be done in C(15,3) ways 3. In a firm are 20 workmen and 10
employees. In how many ways can you have a committee with 3 workmen and 2
employees? First we choose the 3 workmen. This can be done in C(20,3) ways. Then we choose the 2 employees. This can be done in C(10,2) ways. The committee can be assembled in C(20,3).C(10,2) ways 4. In how many ways can you take 5
cards, with at least 2 aces, out of a game of 52 cards? First consider 5 cards, with exactly 2 aces. For the two aces we have C(4,2) possibilities and for the three other cards we have C(48,3) possibilities. The 5 cards can be chosen in C(4,2).C(48,3) ways. Analogous 5 cards, with exactly 3 aces can be chosen in C(4,3).C(48,2) ways. 5 cards, with exactly 4 aces can be chosen in 48 ways. Total = C(4,2).C(48,3) + C(4,3).C(48,2) + 48 5. In how many ways can you split
a group of 13 persons in 3 persons and 10 persons? It is sufficient to choose 3 persons to split the group. This can in C(13,3) ways. 6. How many diagonals are there in
a convex n-polygon? From each angular point we can count (n-3) diagonals and since there are n angular points we have counted n.(n-3) diagonals. But now we have counted twice each diagonal. Hence , there are n(n-3)/2 diagonals. 7. How many numbers consisting of
3 figures, can you make with the figures 0,1,2,3,4 ? There are 4 possibilities for the first figure of the number. There are 5 possibilities for the second figure of the number. There are 5 possibilities for the third figure of the number. Total = 4.5.5 8. How many subset are there in a
set of 10 elements? We can construct a subset of the set by deciding for each of the ten elements, if it belongs to the subset or not. So, for each element there are 2 possibilities. Total possibilities = 210 9. In how many ways can you
arrange m identical stones into k piles so that each pile has at least 1
stone in it. By removing one stone from each pile, this is the number of ways you can arrange m-k identical stones into k (possibly empty) piles. .. Now, view the k piles as a numbered set . Write on each stone the number of a chosen pile and order the stones accordingly. The numbered stones constitute a combination with repetition of k elements (the numbers) choose m-k (the stones). This can be done in
C'(k, m-k) =
C(m-1,m-k) =
10. How many strictly positive
integer solutions ( x, y , z) are there, such that x + y + z = 100? This is the same problem as In how many ways can you arrange 100 identical stones into 3 piles so that each pile has at least 1 stone in it? From previous problem the answer is C(99,97) = 4851 ways 11. How
many terms are contained in (a + b + c)20 All terms can be written as A. ap . bq . cr with p + q + r = 20. The number of terms is the number of solutions of the equation p + q + r = 20 with p, q, r as positive integer unknowns. Now regard (p, q, r) as three ordered elements. Point 20 times one of these elements, and order these elements in the same order as the given elements. This corresponds with one solution of p + q + r = 20 and it is a combination with repetition of 3 elements choose 20. The number of terms is the number of such combinations = C'(3,20) = C(22,20) = C(22,2) = 231 The term A.a10 b3 c7 is contained in (a + b + c)20 . Calculate A. The number of terms a10 b3 c7 is the number of permutations with repetition of the elements a, a, a, a, a, a, a, a, a, a, b, b, b, c, c, c, c, c, c, c This number is 20! ----------- = 22 170 720 10! 3! 7! More problems are available to those who access my
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a3,
a4. In fig (3.2) the order is anti-clockwise, whereas in fig.(3.3) the order
is clockwise. Thus the number of circular permutation of n things which
clockwise and anti-clockwise arrangements give rise to different permutations
is (n-1)! If there are n things and
if the direction is not taken into consideration, the number of circular
permutations is 
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