Answer with explanation:
We have to find ordered pair (a,b) such that , a+b=180
→If there is no constraint on, a and b,that is ,which kind of numbers these are ,there will be infinite number of pairs.
→If we apply the constraint, a≥0, and b≥0 and, a and b are real numbers then also there are infinite number of pairs.
→If we apply the constraint, a≥0, and b≥0 and, a and b are rational numbers then also there are infinite number of pairs.
→But, if a and b are Positive integers ,and taking the constraint, a≥0, and b≥0, the number of pairs are:
Number of integer Points, on the line
a+b=180
1→→If , 0≤a≤90,then value of b will be 180≤b≤90.The pairs will be , {(0,180),(1,179),(2,178),..........(89,91) and (90,90).}
There are 91 pair in all.
2.→→And , if 0≤b≤90,then value of a will be 180≤a≤90.The pairs will be ,{(180,0),(179,1),(178,2),.......(91,89), and (90,90)}.
There are 91 pair in all.
But , the pair (90,90) is included in both the set that is equation 1 and 2 forming pairs.
So, total number of pairs = 91+91-1
=182-1
=181 pair in all.
So,→→ there are 181 pair in all (set of positive integers) having sum 180.
