Question

proved that n(A union B)=n(A)+n(B)-n(A union B )​

Answer 1

To Understand it easily lets say a and b are 2 sets of objects(may be numbers)

aub contains all the objects contained both in a and b taken one at a time

So n(aub) implies the number of objects contained in both a and b taken one at a time.

n(a)=No. of objects in a

n(b)=No. of objects in b

anb contains objects common between sets a and b

Now, n(a)+n(b) contains total number of objects in a and b including those which are common to a as well as b. Which means, those objects are counted twice as they constitute both a and b.

So, to exclude them, such that we count all the objects in a and b only once, we write…

n(a)+n(b)-n(anb)

But the above sentence is the definition of n(aub).

Therefore, n(aub)=n(a)+n(b)-n(anb)

Answer 2

Answer:

To Understand it easily lets say a and b are 2 sets of objects(may be numbers)

aub contains all the objects contained both in a and b taken one at a time

So n(aub) implies the number of objects contained in both a and b taken one at a time.

n(a)=No. of objects in a

n(b)=No. of objects in b

anb contains objects common between sets a and b

Now, n(a)+n(b) contains total number of objects in a and b including those which are common to a as well as b. Which means, those objects are counted twice as they constitute both a and b.

So, to exclude them, such that we count all the objects in a and b only once, we write…

n(a)+n(b)-n(anb)

But the above sentence is the definition of n(aub).

Therefore, n(aub)=n(a)+n(b)-n(anb)