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