We have to find a minimum no. of edges. So for c connected components let's assume c-1 independent vertices which are also c-1 components and remaining 1 component will contain n-(c-1) vertices or n-c+1 vertices. So for minimum no. of edges will be no. of vertices in the single component (which contains n-c+1 vertices) -1, which is (n-c+1)-1 = n-c

∼∀x P(x) = ∃x ∼P(x)

Given data,
Committee members=6,
Mens=6
Women=4
Equal numbers of men and women are= ?
3 men out of 6, 3 women out of 4
⇒ ⁶C₃ * ⁴C₃ ways
⇒ 20 * 4
⇒ 80
Note: There are 3 possibilities to consider an equal number of men and women.
Choose 4 men and 4 women then we get ⁶C⁴ * ⁴C₄ ways = 15
Choose 2 men and 2 women then we get ⁶C₂ * ⁴C^{2 ways = 90
Choose 3 men and 3 women then we get 6C₃ * 4C₃ ways = 80}

Each “boolean” variable has two possible values i.e 0 and 1.
Number of variables= n
Number of input combinations is 2ⁿ.
Each “boolean” function has two possible outputs i.e 0 and 1.
Number of boolean functions possible is 2^{2^n}.
Formula: The number of m-ary functions possible with n k-ary variables is m^{k^n}.

A function is invertible if and only if it is both one-one and onto.

Let A = probability of finding an employee with programming skills
Let B = probability of finding an employee with documentation skill
So A=0.7
B=0.6
AUB=0.9
Hence ,A∩B = A + B - (AUB)
= 0.7+0.6-0.9
= 0.4

Now,
x=((AB)-(BC))
=(b,c)-(e,f))
=b
y=(A-(AC))-(A-B)
=((a,b,d,e)-(d,e))-(a,d)
=(a,b)-(a,d)
=b
Hence, x=y

If a relation is equivalence then it must be
i) Symmetric
ii) Reflexive
iii) Transitive
If a relation is a partial order relation then it must be
i) Reflexive
ii) Anti-symmetric
iii) Transitive
If a relation is total order relation then it must be
i) Reflexive
ii) Symmetric
iii) Transitive
iv) Comparability
Given ordered pairs are (x,y)R(low,up) if (x,up).
Here < , > are using while using these symbols between (x,y) and (y,up) then they are not satisfy the reflexive relation. If they use (x< =low) and (y >=low) then reflexive relation can satisfies.
So, given relation cannot be an Equivalence. Total order relation or partial order relation.

The Hasse diagram representing a lattice, P when turned upside-down represents the poset Q which is dual of P and lattice.

We will use binomial distribution here.
Total possible outcomes will be 6*6=36
Lets first calculate probability of getting sum as 9. We will get sum as 9 when first dice and second dice will have number (3,6) ,(4,5) ,(5,4) ,(6,3). So total 4 possibility are there out of 36.
So Let probability of getting sum as 9 = P(A) = 4/36 = 1/9
Now
Probability of scoring 9 at least in one of the two times
= Probability of scoring 9 exactly one time + Probability of scoring 9 exactly two times

Binomial distribution will be used. And will be calculated as ²⁰C₂ = (0.1)² (0.9)¹⁸

From given description we can conclude that all the elements above the diagonal are 0. So we can say that,
1st row will have 1 non-zero element,
2nd row will have 2 non zero elements,
3rd row will have 3 non-zero elements ,
:
:
:
nth row will contain n non-zero elements.

Therefore total no. of non-zero elements are,
1+2+3+...........+n = n(n+1)/2

Here we will start traversing the expression from left to right and if an operand is encountered then it will be pushed on the top of stack and if an operator is encountered then top two elements will be popped off from the top of stack and current operator is applied on those two operands and then in the end the result of operation will be pushed back on the top of stack. This procedure continues till the whole expression is traversed.


Step: 7
Hence option 4 is the correct answer.

Worst case of deleting a node from a linked list is when we are on the last node of the list and we want to delete (n-1)th node from the list. In this case above lists will have the following time complexity:
Doubly linked list: Since it is doubly linked list so last node will have pointer to its previous node(i.e “n-1” ) node. So (n-1)th node can be deleted in O(1) time. Single Linked List: Since in this no node have a pointer to its previous node so even if we have a pointer on last node of the list we have to traverse (n-1) nodes of the given linked list in order to delete (n-1)th node. Hence its time complexity will be O(n). singly linked circular list: For above scenario since the list is unidirect so we have to traverse (n-1) elements in best case in order to delete (n-1)th node even after having a pointer to nth node.
singly linked list with header node is nothing but Single linked list only. So its time complexity will also be O(n).
Hence Doubly linked list is giving best time complexity to delete a node pointed by ‘p’ from a large collection of nodes constituting the list.

Singly linked circular list with header pointing to the last node is the data structure preferred in the context of concatenating two lists into one list, because it can be done in O(1) time.

Inorder: CABDFEHG
Preorder: DCBAGEFH

Hence, Post-order of above tree will be: ABCFHEGD. Hence, Post-order of above tree will be: ABCFHEGD.

warshall algorithm will take O(n^3) time.
Floyd’s algorithm will take O(n^3) time.
Breadth First traversal will take O(m+n) time.
Depth first traversal will work only if the graph is tree.
Hence among the given options Breadth first traversal is best.

Kruskal’s ,prims , and huffman code generation algorithm require priority queue. But in floyd warshall we use matrix .

Using Primary key indexing all records can be accessed faster.Primary key indexing is based on a key attribute where all the elements of key attribute are arranged in sorted manner.

B-tree and B+ Tree are suitable to organize very large indexed sequential access disk files for interactive random access but since in B+ tree each leaf node have a pointer to its next leaf node then it makes the access of co-sequential batch processing of its records easier.

The best case of insertion sort is when all the elements given in array are already sorted. In that case when insertion sort is applied only one comparison is made for each element of the array. Hence if we have n-elements the n-comparisons will be made. Hence the time complexity will be O(n)

L hospitals Rule, method of back ward substitution and masters theorem is used for time complexity estimation of an algorithm . But Composite trapezoidal rule gives us a technique to approximate the integral on a given interval [a, b] and not used for time complexity estimation of an algorithm.

The QuickHull algorithm is a Divide and Conquer algorithm similar to QuickSort.

Gaussian elimination method is transform and conquer approach.This algorithm solves a system of linear equations by first transforming it to another system with a special property that makes finding a solution quite easy.

NP Problem: Each input to the problem should be associated with a set of solutions of polynomial length, whose validity can be tested in polynomial time. The complexity class of problems of this form is called NP, an abbreviation for "nondeterministic polynomial time".
NP-Hard: A problem is said to be NP-hard if everything in NP can be transformed in polynomial time into it.
A problem is NP-complete if it is both in NP and NP-hard. The NP-complete problems represent the hardest problems in NP.
The list below contains some well-known problems that are NP-complete when expressed as decision problems.
Boolean satisfiability problem (SAT)
Knapsack problem
Hamiltonian path problem
Travelling salesman problem (decision version)
Subgraph isomorphism problem
Subset sum problem
Clique problem
Vertex cover problem
Independent set problem
Dominating set problem
Graph coloring problem

In-Place sorting algorithm is one which requires only O(1) space in order to sort n-elements.
Merge sort has space complexity of O(n). Hence it is not an in-place sorting algorithm.