Given data,
-- 20 men and 12 women
-- 5 people can choose from men and women
-- Each committee contains at least three women
Step-1: We have a constraint that each committee contains atleast 3 women.
possibility-1: 2 men + 3 women.
possibility-2: 1 men + 4 women.
possibility-3: 0 men + 5 women.
Step-1: They are asking to find all possibilities.
= (possibility -1)+ (possibility -2) + (possibility -3)
=​ . ​ (​ ²⁰​ C_{​ 2}.​ *​ ¹²​ C_{​ 3}​ ) + (​ ²⁰​ C_{​ 2}.​ *​ ¹²​ C​ ₂​ ) + (​ ²⁰​ C_{​ 2}.​ *​ ¹²​ C_{​ 2}​ )
= (190*220) + (20*495) + (1*792)
= 41800 + 9900 + 792
= 52492

(a)TRUE: A connected multigraph has an Euler Circuit if and only if each of its vertices has even degree.
(b)TRUE: A connected multigraph has an Euler Path but not an Euler Circuit if and only if it has exactly two vertices of odd degree.
(c)TRUE: A complete graph (K​ _n​ ) has a Hamilton Circuit whenever n ≥ 3. (d) FALSE: A cycle over six vertices (C_{​ 6}​ ) is not a bipartite graph but a complete graph over 3 vertices is bipartite.

(a)TRUE: The set of negative integers is countable.
Suppose negative integers set size is 10.
Ex: -1, -2, -3,.....,-10 is countable
(b)TRUE: The set of integers that are multiples of 7 is countable.
Suppose set of integers size is 10.
Ex: 1*7, 2*7, 3*7, .....,10*7 is countable
(c)TRUE: The set of even integers is countable.
Suppose set of even integers size is 10.
Ex: 2,4,6,8,10,....,20
(d) FALSE: The set of real numbers between 0 and 1⁄2 is countable. We can’t count real numbers.
Ex: 0.1, 0.2, 0.3, .....,0.∞

A bipartite graph (or bigraph) is a graph whose vertices can be divided into two disjoint and independent sets U and V such that every edge connects a vertex in U to one in V. Vertex sets U and V are usually called the parts of the graph. Equivalently, a bipartite graph is a graph that does not contain any odd-length cycles.
→ The two sets U and V may be thought of as a coloring of the graph with two colors.
Option A: FALSE because V​ ₅​ , V_{​ 7}​ and V_{​ 3}​ are adjacent. So, it not not bipartite graph.

Option-B FALSE because V​ ₅​ , V_{​ 6}​ and V_{​ 2}​ are adjacent. So, it not not bipartite graph.

Option-C TRUE because it follows properties of bipartied and no two colours are adjacent.

Option-D FALSE because because V​ ₄​ , V_{​ 6}​ and V_{​ 8}​ are adjacent. So, it not not bipartite graph.

Above all graphs are graceful.

Given P(m,n) ="m divides n"
Statement-A is ∃m ∀n P(m, n). Here, there exists some positive integer which divides every positive integer. It is true because there is positive integer 1 which divides every positive integer.
Statement-B is ∀n P(1, n). Here, 1 divided every positive integer. It is true.
Statement-C is ∀m ∀n P(m, n). Here, every positive integer divided every positive integer. It is false.

→ Vacuous proof is a proof that the implication p → q is true based on the fact that p is false.
→ Trivial proof is a proof that the implication p → q is true based on the fact that q is true.
→ Direct proof is a proof that the implication p → q is true that proceeds by showing that q must be true when p is true.
→ Indirect proof is a proof that the implication p → q is true that proceeds by showing that p must be false when q is false.

An abelian group is a set, A, together with an operation • that combines any two elements a and b to form another element denoted a • b. The symbol • is a general placeholder for a concretely given operation. To qualify as an abelian group, the set and operation, (A, •), must satisfy five requirements known as the abelian group axioms:
Closure: For all a, b in A, the result of the operation a • b is also in A.
Associativity: For all a, b and c in A, the equation (a • b) • c = a • (b • c) holds.
Identity element: There exists an element e in A, such that for all elements a in A, the equation e • a = a • e = a holds.
Inverse element: For each a in A, there exists an element b in A such that a • b = b • a = e, where e is the identity element.
Commutativity: For all a, b in A, a • b = b • a.
A group in which the group operation is not commutative is called a "non-abelian group" or "non-commutative group".

In this problem, we have to find series of evaluation.
Let x=5, inp=5
Iteration-1:
for(i=1; i<=inp; i++)
/* condition true 1<=5 */
{
term = term * x/i;
/* (1*5)/1=5. Here, 5 nothing but ‘x’ */
sum = sum + term ; /* 0+5=5 */
}
Iteration-2: Here, ‘i’ becomes 2.
for(i=2; i<=inp; i++)
/* condition true 2<=5 */
{
term = term * x/i;
/* (5*5)/2 is nothing but x​ 2​ /i */
sum = sum + term ; /* 5+12=17 */
}
;;;
;;;
Iteration-5: Here, ‘i’ becomes 5.
for(i=5; i<=inp; i++)
/* condition true 5<=5 */
{
term = term * x/i;
/* (625*5)/5! is nothing but x​ ⁵​ /i! */
sum = sum + term ;
}
So, Option-B is correct answer.

→ A publicly derived class is a subtype of its base class.
→ Inheritance is defined as deriving new classes (sub classes) from existing ones (super class or base class) and forming them into a hierarchy of classes.
→ Inheritance allows programmers to create classes that are built upon existing classes, to specify a new implementation while maintaining the same behaviors (realizing an interface), to reuse code and to independently extend original software via public classes and interfaces.