Discrete data includes numeric data values that are countable.
Some common examples of discrete data sets include:
Discrete data can be collected through questionnaires, surveys, interviews, or observations.
Once the data has been collected, it can be graphed, which provides a visualization of the data. Pictographs and bar graphs (or bar charts) are common ways to represent discrete data. You can also use histograms, box plots, pie charts, or frequency tables, however note that these can also be used for continuous data.
Besides graphs, data analysis of discrete data may include calculating the mean, median, range or mode. Particularly if the data is being used for statistical analysis.
Finally, the graphs and/or measures of center can be used to answer questions about the data.
For example,
How many cats do the students in my class have?
The data is discrete, because there are a limited number of answers. You can’t have 0.5 of a cat, just like you can’t have 10,000.
Once the data is collected, you can use a bar graph to display it.
Now, you can use the data in the table and the bar graph to answer questions like…
0
cats5
cats2
cats than1
cat?2
students
The Discrete Mathematics questions and answers will help students quickly grasp almost all of the concepts discussed in “Discrete Mathematics.” Students can use these questions to quickly understand the concept. We have also included some practice questions so they can understand better. Furthermore, you can get thorough explanations for every question and cross-check your answers. Click here to read more about discrete mathematics.
Introduction to Discrete Mathematics:
The field of mathematics known as discrete mathematics is concerned with the study of discrete mathematical structure. There are two different types of data: discrete and continuous. Instead of studying continuous data, discrete mathematics examines discrete data. Finite mathematics is another name for it. Discrete mathematics includes some important concepts such as logic, sequences and series, set theory, graph theory, permutation, induction, combinations, etc.
1. If a set B has n elements, then what is the total number of subsets of B. Justify your answer.
Solution:
If a set B has “n” elements, then the total number of subsets of B is 2n.
For example, if B contains 5 elements, say B = {1, 2, 3, 4, 5}, then the total number of subsets of B is 25 = 32.
2. If X and y are the two finite sets, such that n(X U Y) = 36, n(X) = 20, n(Y) = 28, then find n( X ∩ Y).
Solution:
Given: n(X) = 20, n (Y) = 28, n (X U Y) = 36.
As we that, n(X U Y) = n(X) + n(Y) – n(X ∩ Y)
On rearranging the above formula, we get;
n(X ∩ Y) = n(X) + n(Y) – n(X U Y)
Now, substitute the given values in the above formula, we get;
n(X ∩ Y) = 20 + 28 – 36
n(X ∩ Y) = 48 – 36
n(X ∩ Y) = 12.
Hence, n(X ∩ Y) is 12.
3. In a group of 100 students, 72 students can speak English and 43 students can speak Hindi. Based on these data, answer the following questions:
Solution:
Let E be the set of students who speak English, and H be the set of students who speak Hindi.
Thus, from the given conditions, we get
n(E) = 72
n(H) = 43
n(EUH) = 100
Using the formula n(A U B) = n(A) + n(B) – n(A ∩ B), we can write;
n(E U H) = n(E) + n(H) – n(E ∩ H) …(1)
Finding Number of students who can speak both English and Hindi:
Rearranging the formula (1), we get
n(E ∩ H) = n(E) + n(H) – n(E U H)
Now, substitute the known values in the above equation, we get
n(E ∩ H) = 72 + 43 – 100
n(E ∩ H) = 115 – 100
n(E ∩ H) = 15.
Therefore, the number of students who can speak both English and Hindi is 15.
Finding Number of students who speak English only:
n(E – H) = n(E) – n(E ∩ H)
Thus, n(E – H) = 72 – 15
n(E – H) = 57.
Hence, the number of students who speak English only is 57.
Finding Number of students who speak Hindi only:
n(H – E) = n(H) – n(E ∩ H)
n(H – E) = 43 – 15
n(H – E) = 28.
Therefore, the number of students who speak Hindi only is 28.
Also, read: Sets.
4. Find the missing number in the sequence: 5, 10, 15, ___, 25, 30.
Solution:
Given sequence: 5, 10, 15, ___, 25, 30.
In the given sequence, each preceding term is added with 5 to get the succeeding term.
I.e., 5 + 5 = 10
10 + 5 = 15
15 + 5 = 20
20 + 5 = 25
25 + 5 = 30.
Hence, the missing term in the given sequence is 20.
Therefore, the complete sequence is 5, 10, 15, 20, 25, 30.
5. Determine the missing term in the sequence: 4, 12, __, 108, 324, 972.
Solution:
Given Sequence: 4, 12, __, 108, 324, 972.
In the given sequence, each succeeding term is obtained by multiplying 3 to its previous term.
I.e., 4 × 3 = 12
12 × 3 = 36
36 × 3 = 108
108 × 3 = 324
324 × 3 = 972.
Hence, the missing term in the given sequence is 36.
Thus, the complete sequence is 4, 12, 36, 108, 324, 972
Also, read: Sequence and Series.
6. Calculate the tenth term of the Arithmetic Progression 13, 8, 3, -2 …?
Solution:
Given AP: 13, 8, 3, -2…
Here, the first term, a = 13
Common difference, d = -5.
I.e., d = 8 – 13 = -5
3 – 8 = -5
-2 -3 = -5.
And, n = 10.
The formula to find the nth term of Arithmetic Progression is Tn = a + (n-1)d
Now, substitute the given values in the formula, we get
T10 = 13 + (10 – 1)(-5)
T10 = 13 + (9)(-5)
T10 = 13 – 45
T10 = -32
Hence, the tenth term of given Arithmetic Progression 13, 8, 3, -2… is -32.
7. In a group of students, there are 6 boys and 4 girls. Out of 10 students, 4 students have to be selected. Find out how many different ways the students can be selected such that at least one boy should be selected?
Solution:
Total number of students = 10
Number of boys = 6
Number of girls = 4.
So, here we have to select 4 students. In that, at least one boy should be selected.
Hence, there are four cases:
Case 1: All four students are boys:
Thus, the number of ways = 6C4 = 15
Case 2: 3 Boys and 1 girl are selected:
The number of ways to select 3 boys and 1 girl = 6C3 × 4C1 = 20 × 4 = 80.
Case 3: 2 boys and 2 girls are selected
The number of ways to select 2 boys and 2 girls = 6C2 × 4C2 = 15 × 6 = 90.
Case 4: 1 boy and 3 girls are selected
The number of ways to select 1 boy and 3 girls = 6C1 × 4C3 = 6 × 4 = 24.
Therefore, the number of ways to select 4 students such that at least one boy to be selected = 15 + 80 + 90 + 24
= 209.
8. Find out the number of ways that the letters of the word “LEADER” can be arranged?
Solution:
In the word “LEADER”, there are 6 letters.
In that, E is repeated twice.
Hence, the total number of ways that the letters can be arranged = 6!/2!
As we know, 6! = 720 and 2! = 2
Therefore, the number of ways that the letters can be arranged = 720/2 = 360.
9. Find out the number of ways that the letters of the word “DETAIL” can be arranged such that the vowels must occupy odd positions.
Solution:
In the word “DETAIL”, there are 6 letters.
In that, there are 3 vowels (E, A, I), and 3 consonants (D, T, L).
According to the given question, the vowels must occupy odd positions.
So, first, mark the positions of the letters as 1, 2, 3, 4, 5, and 6.
As per the questions, vowels should occupy positions 1, 3, and 5.
Hence, the number of ways to arrange the vowels = 3P3 = 6.
So, the number of ways to arrange the consonants in the remaining positions = 3P3 = 6
Therefore, the number of ways to arrange the letters = 6 × 6 = 36.
Also, read: Permutations and Combinations.
10. Simplify the expression (x + y)(x + z) using the laws of boolean algebra.
Solution:
Let Q = (x + y)(x + z)
Using the distributive law, we can write;
Q = x.x + x.z + y.x + y.z
By applying the idempotent law A.A = A.
Q = x + x.z + y.x + y.z
Q = x(1 + z) + y.x + y.z [Using distributive law]
Applying identity OR law (1 +A = 1), we can write
Q = x. 1 + y.x + y.z
Q = x + y.x + y.z
Again using the distributive law, we get
Q = x. (1 + y) + y.z
Q = x. 1 + y.z (By applying identity OR Law)
Q = x + y.z.
Therefore, the simplification of the expression (x + y)(x + z) is x + y.z.
Solve the following discrete mathematics questions:
2. Find out the number of ways that 3 postcards can be posted in 5 post boxes.
3. Simplify the boolean expression: A = XY + X (Y+Z) + Y(Y+Z).
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