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Discrete Mathematics, also called finite Mathematics or Decision Mathematics, is the study of Mathematical structures that are fundamentally discrete in the sense of not supporting or requiring the notion of continuity. Objects studied in finite Mathematics are largely countable sets such as integers, finite graphs, and formal languages. Discrete Mathematics has become popular in recent decades because of its application in Computer Science.
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Matrices: A rectangular array of entries is called a Matrix. The entries may be real, complex or functions. The entries are also called as the elements of the matrix. The rectangular array of entries are enclosed in an ordinary bracket or in square bracket.
Determinants: Let A = [aij] be a square matrix. We can associate with the square matrix A, a determinant which is formed by exactly the same array of elements of the matrix A. A determinant formed by the same array of elements of the square matrix A is called the determinant of the square matrix A and is denoted by the symbol det. A or |A|.
Combinations: The selection of a number of things taking some or all of them at a time are called combinations.
Sequence: A set of numbers arranged in a definite order according to some definite rule is called a sequence. A sequence is a function whose domain is the set N of natural numbers.
Series: Indicated sum of the terms in a sequence is called a series. The result of performing the additions is the sum of the series.
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Question: In how many ways can 6 men and 5 women be seated in a line so that no two women sit together?
Answer: Let the women be denoted by W and the men be denoted by M.
Now the women can sit at the places marked by '.' .
We have seven places for 5 women. The women can sit in 7P5 ways. Also 6 men can be arranged in 6! ways.
Question: In how many ways can 10 examination papers be arranged so that best and the worst are never together?
Answer: 10 papers can be arranged in 10! ways. When the best and the worst are together, the number of arrangements will be 9! x 2.
When the best and the worst are not together, the number of arrangements will be 10! - (9!)2 = 9!(10 - 2) = (8) x (9!).
Question: In how many ways can three prizes be distributed among 4 boys when
i) No one gets more than one prize.
ii) A boy can get any number of prizes.
Answer: i) The first prize can be given in 4 ways as one cannot get more than one prize, the remaining two prizes can be given in 3 and 2 ways respectively.
The total number of ways = 4 x 3 x 2 = 24.
ii) As there is no restriction, each prize can be given in 4 ways.
The total number of ways = 43 = 64.
Question: From a class of 32 students, 4 are to be chosen for a competition. In how many ways can this be done?
Answer: We are to select 4 students from 32.
This selection can done in
Question: A sports team of 11 students is to be constituted choosing at least 5 from class XI and 5 at least from XII. If there are 20 students in each of these classes, in how many ways can the team be constituted?
Answer: Number of students in each class is 20.
Total number of selections = 2 x 600935040 = 1201870080
Question: Given five different green dyes, four different blue dyes and three different red dyes, how many combinations of dyes can be chosen taking atleast one green and one blue dye?
Answer: The least number of dyes that a combination can have is 2.
(one blue and one green).
Maximum number of dyes that a combination can have is 12
(5G, 4B, 6R).
Atleast one green dye can be selected out of 5 green dyes. The number of ways is
After selecting one or more green dyes, we can select atleast one blue dye out of 4 different blue dyes. The number of ways is
After selecting atleast one green dye and atleast one blue dye, at least one red dye or no red dye can be selected in
= 1 + 3 + 3 + 1 = 8 ways
The total number of combinations (31)(15)(8) = 3720.
Question: A letter lock consists of three rings, each marked with fifteen different letters. Find in how many ways it is possible to make an unsuccessful attempt to open the lock.
Answer: 1st ring can be attempted in 15 ways.
1st and 2nd ring can be attempted in 15 x 15 = 152 ways
1st and 2nd and 3d ring can be attempted in 15 x 15 x 15 = 153 ways
Among these 153 attempts, one attempt will be a successful attempt.
Hence the number of unsuccessful attempts is 153-1= 3375 -1 = 3374
Question: Find the sum of all four digit numbers formed by using 2, 3, 6, 9 in which no digit is repeated.
Answer: If 2 occupies unit's place, the remaining 3 digits can be arranged in 3!= 6ways. Similarly, if 2 occupies ten's place, hundred's place, thousand's place, in each of these cases we get 3! numbers. Thus, the positional value contributed by 2 to the sum when it occupies different
(3!)(2) + (3!)(20) + (3!)(200) + (3!)(2000) = 3!(2)(1111)
Similarly, the values contributed by 3, 6, 9 to the sum are
3! (3)(1111), 3! (6) (1111), 3! (9)(1111) respectively.
The required sum is 3!(1111)(2 + 3 + 6 + 9) = 1,33,320
|More topics in Discrete Math|
|Mathematical Logic||Set Theory|
|Relations and Functions||Difference between Relation and Function|
|Sets and Relations||Sequences and Series|
|Boolean Laws||Mathematical Induction|
|Graph Theory||Pigeonhole principle|
|Partially ordered set|