IGNOU MCS-013-Discrete Mathematics, Latest Solved Assignment (July 2023 - January 2024 )
Q4. (a) How many words can be formed using letter of “EXCELLENT” using each letter at most once?
I) If each letter must be used,
Sol) To find the number of words that can be formed using the letters of the word”EXCELLENT” with each letter used at most once, we can use the concept of permutations.
The word “EXCELLENT” has a total of 9 letters, with repetitions. Since you want to use each letter at most once, you are essentially asking for the number of permutations of these 9 letters.
The formula to calculate permutations of n items taken r at a time (where n is the total number of items and r is the number of items we want to arrange) is given by:
P(n,r) = n! / (n-r)!
In this case, n=9 (total number of letters in “EXCELLENT”) and r=9 (using all letters). Plugging these values into the formula:
P (9,9) = 9! / (9-9)!
=9!
=362,880
So, there are 362,880 words that can be formed using the letters of “EXCELLENT” with each letter used at most once.
ii) Some or All Letters May Be Omitted:
If some or all of the letters can be omitted, then for each letter in “EXCELLENT,” we have two choices: include it in the word or omit it. This leads to a total of possible combinations, as each letter can either be included or omitted independently.
So, there are 512 words that can be formed using the letters of “EXCELLENT” where some or all of the letters may be omitted.
2^9 = 512.
So, there are 512 words that can be formed using the letters of “EXCELLENT” where some or all of the letters may be omitted.
(b) What is a relation? What are different types of relation? Explain equivalence relation with the help of example.
Sol) In mathematics, a relation is a set of ordered pairs that describes the connection between elements from two different sets. It establishes a certain connection or association between elements, which can represent various types of interactions or dependencies.
There are several types of relations, each with its own characteristics:
Binary Relation: A relation between two sets A and B is a subset of the Cartesian product A×B, where each element of the relation is an ordered pair (a, b) where a is from set A and b is from set B.
Reflexive Relation: A relation on a set where every element is related to itself. In other words, for all a in the set, (a, a) is in the relation.
Symmetric Relation: A relation on a set where if (a, b) is in the relation, then(b, a) is also in the relation.
Transitive Relation: A relation on a set where if (a, b) and (b, c) are in the relation, then (a, c) is also in the relation.
Equivalence Relation: An equivalence relation is a relation that is reflexive, symmetric, and transitive. It divides the set into distinct groups or equivalence classes, where elements within the same class are considered equivalent.
Let’s explain equivalence relation with an example:
Example: Equivalence Relation on Integers
Consider the set of integers Z. Define a relation R such that two integers a and b are related by R if their difference is a multiple of 3. Formally, aRb if a−b is divisible by 3.
Reflexive: For any integer a−a=0, which is divisible by 3. So, aRa for all a in Z.
Symmetric: If aRb, then a−b is divisible by 3. This implies that b−a is also divisible by 3, and thus bRa.
Transitive: If aRb and bRc, then a−b and b−c are both divisible by 3. This implies that a−c=(a−b)+(b−c) is also divisible by 3, so aRc.
The relation R in this example is an equivalence relation because it satisfies all three properties: reflexivity, symmetry, and transitivity. It partitions the integers into equivalence classes based on their remainders when divided by 3. For example, the equivalence class of 0 contains all integers divisible by 3, the equivalence class of 1 contains integers with a remainder of 1 when divided by3, and so on.
(c) Prove that 1 2 +2 2+3 2+ …+ n 2 = n(n+1)(2n+1)/6 ; n ∈N
Sol) Step 1: Base Case
First, we prove the formula for the base case n=1
1(1+1)(2(1)+1)/6
=1
The base case holds true
Step 2: Inductive Hypothesis
Assume that the formula holds true for n=k, where k is an arbitrary positive integer. In other words, assume that
Step 3: Inductive Step
Now, we need to prove that the formula holds for n=k+1:
We want to show that
12+2 2+3 2+…..k 2+(k+1)2=(k+1)(k+2)(2k+3)/6
LHS:
=k(k+1)(2k+1)/6 + (k+1) 2
=(k(k+1)(2k+1)+ 6(k+1) 2 )/6
=(k+1)(k(2k+1)+6(k+1))/6
=(k+1)(2k 2+7k+6)/6
=(k+1)(k+2)(2k+3)/6
Since the inductive step holds, the formula is proven by mathematical induction for all positive integers n.
(d) What is counterexample? Explain its use with the help of an example
Sol) A counterexample is a specific example that contradicts or disproves a given statement, theory, or hypothesis. It’s used to show that a general claim is not universally true by providing a specific case where the claim fails to hold.
In other words, a counterexample is used to demonstrate that a statement is false by presenting a scenario that goes against the claim being made
Example: Even Numbers and Squares
Claim: “All even numbers are perfect squares.”
Even number: 10
Perfect squares: 1, 4, 9, 16, …
Since 10 is not a perfect square, it serves as a counterexample to the claim that “All even numbers are perfect squares.” This counterexample demonstrates that the statement is false, as it provides a specific instance that contradicts the claim
The role of a counterexample is to challenge the validity of a general statement by highlighting its limitations or exceptions. It shows that the statement doesn’t hold true in all cases, thus prompting a revaluation of the claim or hypothesis
Q5. (a) How many different professionals committees of 8 people can be formed, each containing at least 2 Doctors, at least 2 Public Servants and1ITExpert from list of 7 Doctors, 6 Public Servants and 6 IT Experts?
Sol) We have:
7 Doctors
6 Public Servants
6 IT Experts
Step 1: Selecting Doctors
Since each committee must have at least 2 Doctors, we have two scenarios:
(7
2 ) ways to select 2 Doctors from 7.
(18
6 ) ways to select 6 members from the remaining 18 (Public Servants +ITExperts).
Total ways for this scenario:
( 7 x ( 18
2 ) 6 )
Scenario 2: Selecting 3 Doctors and 5 additional members.
(7
3) ways to select 3 Doctors from 7.
(
18
5 ) ways to select 5 members from the remaining 18 (Public Servants +ITExperts).
Total ways for this scenario:
(7 x (18
3 ) 5)
Step 2: Selecting Public Servants
Since each committee must have at least 2 Public Servants, we have two scenarios:
Selecting 2 Public Servants and 4 additional members.
Selecting 3 Public Servants and 3 additional members.
Scenario 1: Selecting 2 Public Servants and 4 additional members.
(6
2) ways to select 2 Public Servants from 6.
(18
4 )ways to select 4 members from the remaining 18 (Doctors + IT Experts).
Total ways for this scenario:
(6 x (18
3) 3)
Sum up the total ways from all scenarios:
Total ways = Total ways from Doctors scenarios + Total ways from Public Servants scenarios + Total ways from IT Experts scenarios
Finally, add up the results from all scenarios to get the total number of different professional committees that can be formed while meeting the given conditions.
(b) A and B are mutually exclusive events such that P(A) = 1/2 and P(B) =1/3and P (AU B) = 1/4. What is the probability of P(A Ո B)?
Sol) Mutually exclusive events are events that cannot occur at the same time. In this case, if events A and B are mutually exclusive, then the probability of their union (A ∪ B) is the sum of their individual probabilities: P(A ∪B) = P(A) +P(B).
Given the information:
P(A) = 1/2
P(B) = 1/3
P(A ∪ B) = 1/4
However, it’s important to note that if events A and B are mutually exclusive, their union is essentially the sum of their probabilities, which cannot exceed1. This is because they cannot both occur at the same time, so their combined probability cannot be greater than the probability of any single event.
But in this case, P(A) + P(B) = 1/2 + 1/3 = 5/6, which is greater than 1. This means that events A and B cannot be mutually exclusive. Therefore, there seems to be an inconsistency in the given information.
(c) Find how many 3 digit numbers are odd?
Sol) A three-digit number is odd if its units digit is odd. The odd digits are 1, 3, 5, 7, and 9
To count the number of three-digit odd numbers, we have several possibilities for the hundreds and tens digits:
Hundreds digit is 1: In this case, the tens digit can be any of the odd digits (1, 3, 5, 7, or 9), and the units digit can also be any of the odd digits. So, there are 5 choices for the tens digit and 5 choices for the units digit, resulting in 5×5 =25 possibilities.
Hundreds digit is 2, 3, 4, 5, 6, 7, or 8: In this case, the tens digit can again be any of the odd digits (1, 3, 5, 7, or 9), and the units digit can also be any of the odd digits. So, there are 5 choices for the tens digit and 5 choices for the units digit for each of the seven choices for the hundreds digit, resulting in 7×5×5=175 possibilities.
Hundreds digit is 9: This case is similar to the first case, with 5 choices for the tens digit and 5 choices for the units digit.
Adding up the possibilities from all cases:
25+175+25=225
So, there are 225 three-digit odd numbers.
(d) Explain whether the function f(x) = x + 1 is one-one or not.
Sol) A function is said to be one-to-one (or injective) if each distinct element in the domain is mapped to a distinct element in the codomain. In other words, not two different inputs are mapped to the same output.
For the function f(x)=x+1, let’s examine whether it is one-to-one or not. Assume f(x1)=f(x2) in the domain.
Then,
x1+ 1= x2 + 1
Simplifying, we get x1=x2
Since each distinct input maps to a distinct output, the function f(x)=x+1 is indeed one-to-one.
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