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# Discrete Mathematics Help

Discrete Mathematics is the study of Mathematical structures dealing with objects that can distinct separate values it is also called finite Mathematics or Decision Mathematics. It 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. It has become popular in recent decades due to its application in Computer Science. It is used in software developments, cryptography, programming language, algorithms etc. It covers set theory, graph theory, logic, permutation and combination as well.

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## Discrete Math Online Tutoring

Through online tutoring and homework help offered by TutorVista, students can improve their  grades and boost up their confidence. We help you to solve problems based on Set theory, Graph theory, Number theory, Logic, Permutations and Combinations with ease. Discrete Mathematics is the single most important field of Math hence our excellent online tutors have expertise in all the relevant Math topics. The topics included in the study of discrete math are mentioned below:

Set Theory:
Set Theory is defined as the study of sets which are collection of objects arranged in a group. The set of numbers or objects can be denoted by the braces {} symbol.

For example, the set of first 4 even numbers is {2,4,6,8}

Graph Theory: It is the study of the Graph. Graph is a mathematical structure used to pair  the relation between objects. Graphs are one of the prime objects of study in Discrete Mathematics.

Logic: Logic in mathematics can be defined as the study of valid reasoning. There are three types of logic gates, AND($\wedge$), NOT(~), and OR($\vee$)

Permutation: The different arrangements that can be made with a given number of set taking  some or all of them in a particular sequence at a time are called Permutation. For example, there are six permutation of the set {5,6,7}, namely (5,6,7), (5,7,6), (6,5,7), (6,7,5), (7,5,6), and (7,6,5).

Combination: The selection of a number of objects taking some or all of them at a time is called Combination. The Order of selection does not matter for the Combination.

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 countable set of natural numbers.

Series: A Series is the sum of the terms of a Sequence. The result of adding all the terms together: s1+s2+s3+s4... is the sum of the series.

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## Discrete Math Problems

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.

,$M_1$, $M_2$, $M_3$, $M_4$, $M_5$, $M_6$,

Now the women can sit at the places marked by '.' .
We have seven places for 5 women. The women can sit in $^{7}\textrm{P}_{5}$ ways. Also 6 men can be arranged in 6! ways.

$\therefore$ Total number of arrangements = 6! $\times$ $^{7}\textrm{P}_{5}$ = 1814400

Question: 2012 was a leap year. create a truth table for negation.

Solution: 2012 was not a leap

Truth table for NOT

 Input p Output $\widetilde{p}$ T F F T
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 $^{32}\textrm{C}_{4}$ = $\frac{32.31.30.29}{4.3.2.1}$ = $35963$

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.

$\therefore$ 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 at least 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).

At least one green dye can be selected out of 5 green dyes. The number of ways is

= $^{5}\textrm{C}_{1}$ + $^{5}\textrm{C}_{2}$ + $^{5}\textrm{C}_{3}$ + $^{5}\textrm{C}_{4}$ + $^{5}\textrm{C}_{5}$ ways

= 5 + 10 + 10 + 5 + 1 = 31

After selecting one or more green dyes, we can select at least one blue dye out of 4 different blue dyes. The number of ways is

= $^{4}\textrm{C}_{1}$ + $^{4}\textrm{C}_{2}$ + $^{4}\textrm{C}_{3}$ + $^{4}\textrm{C}_{4}$ = 4 + 6 + 4 + 1 =15

After selecting at least one green dye and at least one blue dye, at least one red dye or no red dye can be selected in

= $^{3}\textrm{C}_{0}$ + $^{3}\textrm{C}_{1}$ $^{3}\textrm{C}_{2}$ + $^{3}\textrm{C}_{3}$ ways

= 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
values is

(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