Question

let A and B are two sets such that n(A)=4,n(B)=7 ,then find the maximum number of elements in A union B​

Answer 1

Answer:

Hello,

Step-by-step explanation:

n(A△B)=n(A∪B)−n(A∩B)

for n (A \triangle B)tobemax.n (A \cap B)=0$$

We know, that  

n(A∪B)=n(A)+n(B)−n(A∩B)=15+25−0=40

⇒n(A△B)  

max

​  

=40−0=40

For minimum value of n(A△B)

n(A∪B) should be min, n(A∩B) should be max.

n(A△B) min =25−15=10

So. value of  

n(A△B)=n(A∪B)−n(A∩B) lies om the set

10,11,12,......,3,9,40

Now, when n(A△B) is max. i.e. when  

n(A∪B)=40 & n(A∩B)=0

If we decrease n(A∪B) by 1 then n(A∩B)

Will increase by 1

n(A△B)=39−1=38

Similarly on for the decrease of 1 you will get in (A△B) as 36 and 30 so on.

Hence  

Range of n(A△B)=10,12,14,16,18,20,......,38,40  

=16 values

n(A) = 4

n(B) = 7

n(AUB) = n(A) + n(B) - n(AnB)

let n(AnB) = 0

n(AUB) = 4 + 7

= 11

Hence:AUB=11

Thank You:)

Answer 2

Answer:

11

Step-by-step explanation:

n(A) = 4

n(B) = 7

n(AUB) = n(A) + n(B) - n(AnB)

let n(AnB) = 0

then

n(AUB) = 4 + 7

= 11

Hence , the max no. of element in AUB is 11

Note:- 'n' in bracket with A and B is intersection symbol..!!

I hope it is helpful for you..!!

Thank you!!