# Engineering Mathematics

Engineering Mathematics
Question:-
Option (A)
2
Option (B)
30
Option (C)
56
Option (D)
256
Correct Option:
C
Question Solution:
This correspond to an ordered partition of 8 elements into two groups the first with 5 elements and seconds with 3 elements. The number of ways of doing this is
P(8;5,3) = (8!/5!3!) = 56
Question:-
Option (A)
k and n
Option (B)
k - 1 and k + 1
Option (C)
k – 1 and n – 1
Option (D)
k + 1 and n – k
Correct Option:
C
Question Solution:
Maximum components will result after removal of node if graph G is a star graph as shown below

Or null graph of n vertices as shown below

In either case, if node v is removed, the number of components will be n-1, where n is the total number of nodes in the star graph
\ n-1 is the maximum number of components possible. Maximum components will result if the node being removed is a lone vertex in which case, the number of components will be k-1
\ The number of components must necessarily lie between k-1 and n-1
Question:-
Option (A)
(2n/n)*2n
Option (B)
3n
Option (C)
(2n)1 / 2n
Option (D)
2n/n
Correct Option:
B
Question Solution:
For each of the n couples invited to the party one of three thing is possible
1. Both husband and wife attend party
2. Wife only attends the party
3. Neither husband nor wife attends the party
Since there are n such couples, total number of possibilities = 3n
Question:-
Option (A)
1/4, 1/2
Option (B)
1/2, 1/4
Option (C)
1/2, 1
Option (D)
1, 1/2
Correct Option:
D
Question Solution:
Given P(A) = 1
P(B) = 1/2
Both events are independent
So, P(A Ç B) = P(A).P(B) = 1.1/2 = 1/2
P(A|B) = {P(AÇB) / P(B)} = {(1/2)/(1/2)} = 1
P(B|A) = {(P(A Ç B)/P(A)} = ((1/2)/1) = 1/2
Question:-
Option (A)
P(x) = True for all x ÎS such that x Î such that x ¹ b
Option (B)
P(x) = False for all x ÎS such that x ¹ a and x ¹ c
Option (C)
P(x) = False for all x ÎS such that b £ x such that x ¹ c
Option (D)
P(x) = False for all x ÎS such that a £ x and b £ x
Correct Option:
D
Question Solution:
If a £ x, since p(x) Þ p(y) whenever x £ y
\ p(a) Þ p(x)
Now since p(a) = True, p(x) = cannot be false
\ (d) cannot be true
Question:-
Option (A)
("x)[a] Þ ("x)[ b] Þ ("x)[a Þ b]
Option (B)
("x)[a]Þ[\$x][a^b]
Option (C)
("x)[aÚb]Þ[\$x][a]Þ("x)[a]
Option (D)
("x)[a Þ b]Þ(("x)[a]Þ ("x)[b])
Correct Option:
D
Question Solution:
("x)[a Þ b] Þ (("x)[a] Þ "(x)[b]) is logical equivalence and therefore, a valid first order formula