Laws of Boolean Algebra Annulment law

Running Head: Discrete Maths 1
Discrete Mathematics
Name of Student
Name of Professor
Date

Discrete Maths 2
Question 1
a) nCr= n!
r ! × ( n−r ) ! = 6 !
1! × ( 6−1 ) ! =6
b) 6
c) 24 =16
d) Choosing bowlers 4 out of 5
Choosing 1 wicketkeeper out of 2
Choosing the remaining 6 players out of 11
( 5 C 4 ) ( 2C 1 ) ( 10 C 6 )+ (5 C 4 ) ( 2 C 2 ) ( 10C 5 ) + ( 5C 5 ) ( 2C 1 ) ( 10 C 5 ) + ( 5 C 5 ) ( 2C 2 ) ( 10 C 4 ) = ( 5× 2× 210 ) + ( 5 ×
¿ 2100+1260+504 +210
¿ 4074
Question 2
Laws of Boolean Algebra
a) Annulment law
A .0=0 ;
A 0
0
A+1=1 ;

Discrete Maths 3
A 1
1
b) Identity law
A+0= A ;
A A
0
A .1= A;
A A
1
c) Indempotent law
A+ A= A;
A A
A
A . A= A;