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.

Get discrete math help in TutorVista and enhance your knowledge in Mathematics. Students can join our online tutoring programs and get the required help from an expert tutor 24/7. Learn basics of discrete math and get online help anytime at the convenience of your home. Our personalized one-on-one tutoring sessions are useful for students wherein our tutors go through extensive training, and keep themselves acquainted with the current requirements measured by the educational boards. Moreover, we follow same textbooks to maintain the academic standard. Our online sessions are brilliantly structured that are relevant to topics of a student’s academic requirement. Experience our useful mathematics help today.

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}

You can get solved problems, answers and worksheets from TutorVista. Make learning Discrete Mathematics simple by choosing tutoring sessions with TutorVista. Some like to gobble up Math while others would rather wane than solve a problem. TutorVista has a comprehensive list of topics in math that experts cover across all grades, get a free demo from our expert math tutors now we are available 24/7.

**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 |

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 = 4^{3} = 64.

**Question**: From a class of 32 students, 4 are to be chosen for a competition. In how many ways can this be done?

**Answe****r**: 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

Given below are some of the practice problems:

Related Topics | |

Math Help Online | Online Math Tutor |