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# Discrete Mathematics Questions and Answers

Discrete Mathematics Question Bank with Answer:
Discrete maths are maintained as a part of mathematics subject, which is essential for any beginner of studies. Discrete maths cover the topic like number system, element of set theory, algebraic equation relation and functions,Probability, graph theory, and matrix. The question bank is prepared based on the availability  and application of math topic. Required number of questions are given with key answers. By practicing Discrete mathematics question and answers makes the learner easy and at the same time enjoy towards the subject.

1. Find the value of 5.123x10-2 + 6.042x10-2

Solution: 5.123 + 6.042 = 11.165x10-2
the value can be divided by 10 which equals to 1.1165x10-1

2. Insert the missing number 5,20,10,_,20,80

Solution: 5,20,10,_,20,80
Here first digit is multiplied by 4,for the obtained number (second digit) divided by 2,these is how the series follows.
5 x 4 = 20
20 ÷ 2 = 10
10 x 4 = 40
40 ÷ 2 = 20
20 x 4 = 80
the missing number is 40.

3. Evaluate 7P5 using formula

Solution: Formula for number of permutation = nPr = $\frac{n!}{\left ( n-r \right )!}$
when n= 7 and r = 5, assign the value in the given formula.

7P5 = $\frac{7!}{\left ( 7-5 \right )!}=\frac{7!}{(2)!}=\frac{7*6*5*4*3*2*1}{2*1}$ = 7*6*5*4*3 = 2520

4. Find the number of combinations of the letter of the word “KEYBOARD” when 5 letter is taken at a time.

Solution: The word “KEYBOARD” has 8   different letters, when 5 letter are taken at a time

The number of combination = ncr = $\frac{n!}{\left ( n-r \right )!r!}$
here n = 8 r = 5             nCr = $\frac{8!}{\left ( 8-4 \right )!4!}$
= $\frac{8!}{(4)!4!}=\frac{8*7*6*5*4*3*2*1}{(4*3*2*1)4*3*2*1}$ = 70

5. Find the value of n when the first term of the series is 10, their common difference is 3 and the nth term Tn = 64

Solution: Given the first term a = 10, d = 3, Tn = 64
The general form the series is Tn = a +(n - 1)d
assign the value of the term here
Tn = a + (n - 1)d
64 = 10 + (n-1)3
64 = 10 + 3n -3
64 = 7 + 3n   (add -7 on side of the equation)
64 - 7 = 7 + 3n - 7
57 = 3n
$\frac{57}{3}$ = n
19 = n

6. Find the sum of the series of geometric progression term 3+6+12+.......to 10 terms

Solution: Given a = 2,  n = 10
r = $\frac{T_{2}}{T_{1}}=\frac{6}{3}$ = 2

= $\frac{T_{3}}{T_{2}}=\frac{12}{6}$ = 2

since r is greater than 1 we have sum of series of term formula  Sn = $\frac{a\left ( r^{n}-1 \right )}{\left ( r-1 \right )}$

S10 = $\frac{2\left ( 2^{10}-1 \right )}{\left ( 2-1 \right )}$

S10 = $\frac{2\left ( 1024-1 \right )}{\left ( 1 \right )}$

S10 = 2 (1023)
S10 = 2046

From Encyclopedia

Mathematics

Question:Hello, I need to solve the discrete mathematics problem, proposition for p and q ................._____......_____ ........................_............_ p <=> q ( p q ) ( q p ) Thanks for your help in advance, B/R. Hey! I would appricate if someone help me :( B/R

Answers:p <=> q ( p q ) ( q p ) p <=> q p q q p p <=> q p q

Question:Ms Pezzulo teaches geometry and then biology to a class of 12 advanced students in a classroom that has only 12 desks. In how many ways can she assign the students to these desks so that (a)no student is seated at the same desk for both classes? (b)there are exactly six students each of whom occupies the same desk for both classes?

Answers:The only feasible way for one to occupy a desk without being "seated" is to stand on the desk. So in the first class session, Group A (consisting of 6 students) sits in their respective chairs and so does Group B (6 students). In session 2 Group A stands in their same desk instead of sitting ang Group B switches seats with one another, like musical chairs. Since this arrangement only involves the movement of Group B, there are 216 (6 students times 6 chairs times 6 chairs) possible seating arrangments for Group B for every 1 seating arrangement for Group A. There are 36 possible seating arrangements for Group A (6 students times 6 chairs). So we have 216*36=7776 So Ms Pezzulo has a whopping 7776 seating arrangment options for our single sit -down/stand-up solution. If you want to be technical with it, you should multilply this solution by thenumber of possible bodily positions one could use to occupy a desk without actually putting one's bottom in the seat. LOL. Have at it! I'm going to bed. X-) ZZZZZZZZZ

Question:I am having some difficulty in my Discrete Mathematics course. I don't want to post the question up here just to get the answer. I am looking for help on understanding the process of how to solve the problem so that I can finish the assignment. Here is the problem: Show that 6 divides (a^3 - a) for all integer numbers a>=2. (Hint: One of 3 sequential integers is divisible by 3). I tried writing out the first few terms but I am still lost on how to attack this problem. Are any websites that can explain discrete mathematics in a simpler manner? Also this question if from the professor himself (he decides to give us problems that are not in the book....so the book is little help)

Answers:The key is noticing that: a^3 a = a(a^2 1) = = a(a 1)(a + 1) But a, a 1 and a + 1 are three consecutive integers, so at least one is even, which means that 2|(a^3 a). Now, if you divide these three integers by 3, you'll see that at least one of the reaminders must be 0, which is to say that the corresponding integer is divisable by 3, so 3|(a^3 a). But gcd(2,3) = 1 so, as a^3 a is divisible by both 2 and 3, it must be divisible by their product, 6.