Step-1: Given number of equivalence classes with 5 elements with three elements in each class will be 1,2,2 (or) 2,1,2 (or) 2,2,1 and 3,1,1.
Step-2: The number of combinations for three equivalence classes are
2,2,1 chosen in (​ ^5​ C_{​ 2}​ *^{​ 3}​ C​ ₂​ *^{​ 1}​ C_{​ 1}​ )/2! = 15
3,1,1 chosen in(​ ^5​ C_{​ 2}​ *^{​ 3}​ C​ ₂​ *^{​ 1}​ C_{​ 1}​ )/2! = 10
Step-3: Total differential classes are 15+10
=25.

Step-1: Given complete graph K​ 4​ .To find total number of spanning tree in complete graph using standard formula is n​ ^(n-2) Here, n=4
=n​ ^(n-2)
= 4​ ²
=16
Step-2: Given Bipartite graph K​ _2,2​ . To find number of spanning tree in a bipartite graph K​ _m,n​ having standard formula is m​ ^(n-1)​ * n^{​ (m-1)}​ .
m=2 and n=2
= 2​ ^(2-1)​ * 2^{​ (2-1)}
= 2 * 2
= 4

A binary relation R over a set X is reflexive if every element of X is related to itself. Formally, this may be written ​ ∀ ​ x ∈X : xRx.
Ex: Let set A={0,1}
R​ ₁​ ={(0,0),(1,1)} all diagonal elements we are considering for reflexive relation.
R​ _2​ ={(0,0),(1,1)} all diagonal elements we are considering for reflexive relation.
R​ ₁​ ∩ R_{​ 2}​ must have {(0,0),(1,1)} is reflexive.
R​ ₁​ ∪ R_{​ 2}​ must have {(0,0),(1,1)} is reflexive.

Given data,
-- 3 cards in a box
-- 1​ st​ card: Both sides of one card is black. The card having 2 sides. We can write it as BB.
-- 2​ nd​ card: Both sides of one card is red. The card having 2 sides. We can write it as RR.
-- 3rd card: one black side and one red side.​ ​ We can write it as BR.
Step-1: The probability that the opposite side is the same colour as the one side we observed is 2⁄3 because total number of cards are 3

Definition of clique is already given in question.
Definition: A clique in a simple undirected graph is a complete subgraph that is not contained in any larger complete subgraph.
Step-1: b,c,e,f is complete graph.

Step-2: ‘a’ is not connected to ‘e’ and ‘b’ is not connected to ‘d’. So, it is not complete graph.

Here, ⊕ is nothing but Ex-OR operator. The truth table for Ex-OR is

According to truth table,
Option-A is TRUE
Option-B is a 1 ⊕ 1 is 0.
0 ⊕ 1 is 1(TRUE)
Option-C is 1 ⊕ 1 is 0.
0 ⊕ 0 = 0 but given 1. So, FALSE
Option-D is TRUE.

→ The IEEE-754 double-precision format to represent floating point numbers has a length of 64 bits
→ In the IEEE 754-2008 standard, the 64-bit base-2 format is officially referred to as binary64 called double in IEEE 754-1985.
→ IEEE 754 specifies additional floating-point formats, including 32-bit base-2 single precision and, more recently, base-10 representations.

Method-1: Using K-Map

Method-2: Using boolean simplification
= x’y’z+x’yz+xy’z’+xyz’
= x'z(y'+y)+ xz'(y'+y)
= x'z+xz' (Since y'+y=1)

(AB’ + AB’C + AC) (A’C’ + B’)
= (AB'+AC) (A'C'+B')
= AB'A'C' + AB'B' + ACA'C' + ACB'
= AB'B' + ACB'
= AB'(C+1)
= AB'

Positive-edge-triggered JK flip-flop is

The Truth Table for the JK Function

When J = 1 and K = 1 , The output continuously Toggles from 1 to 0 and 0 to 1. At the end Output is indeterminate. This condition is called as Race around Condition. This happens when Propagation Delay is less than the Pulse width.

x * y < i + j || k
Step-1: Evaluate x * y because multiplication has more priority than remaining operators
x * y→ 0
Step-2: i + j is 1
Step-3: (x*y) < (i+j) is 1. Because relational operators only return 1(TRUE) or 0(FALSE).
Step-4: ((x*y) < (i+j)) || k is logical OR operator.
1 || -1 will returns 1.
Note: The precedence is ((x*y) < (i+j)) || k

int (*f())[ ] declare a function returning a pointer to an array of integers.