Question

sets and relations. plz plz plz dont make spams​

Answer 1

Question

In a school, on the Republic Day, three dramas A, B and C are performed on the dais. In a group of people, who attended the function and who like at least one of the three drama, 16 people like A, 20 people like B, 15 people like C, 4 people like both A and B, 3 people like both A and C, 3 people like both B and C and 2 people like all the three. Then how many people like at most 2?

Answer

41

Explanation

Here, we have

n(A) = 16

n(B) = 20

n(C) = 15

$\sf n(A\cap B) = 4$

$\sf n(A\cap C) = 3$

$\sf n(B\cap C) = 3$

$\sf n(A\cap B\cap C) = 2$

We are asked for people who like at most 2.

This means that we are asked for people who like either a single group or two groups but not three groups.

This means, that we are asked for people who don't like all three groups.

In term of sets, it can be represented as

$\sf n(A\cup B\cup C) - n(A\cap B\cap C)$

Now, we know that

$\sf n(A\cup B\cup C) = n(A) + n(B) + n(C)\\ - n(A\cap B) - n(A\cap C) - n(B\cap C) +\\ n(A\cap B\cap C)$

So we get

$\sf n(A\cup B\cup C) = 16 + 20 + 15 - 4 - 3 - 3 + 2$

$\sf n(A\cup B\cup C) = 43$

So, $\sf n(A\cup B\cup C) - n(A\cap B\cap C) = 43 - 2 = 41$

Hence, the number of people who like at most 2 is 41.

Answer 2

n(A) = 16

n(B) = 20

n(C) = 15

$\sf n(A\cap B) = 4n(A∩B)=4</p><p></p><p>\sf n(A\cap C) = 3n(A∩C)=3 \\ \\ </p><p></p><p>\sf n(B\cap C) = 3n(B∩C)=3 \\ \\ </p><p></p><p>\sf n(A\cap B\cap C) = 2n(A∩B∩C)=2$

$n(A∪B∪C)=n(A)+n(B)+n(C)</p><p>−n(A∩B)−n(A∩C)−n(B∩C)+</p><p>n(A∩B∩C)$

$\sf n(A\cup B\cup C) = 16 + 20 + 15 - 4 - 3 - 3 + 2n(A∪B∪C)=16+20+15−4−3−3+2 \\ \\</p><p></p><p>\sf n(A\cup B\cup C) = 43n(A∪B∪C)=43 \\ \\</p><p></p><p>So, \sf n(A\cup B\cup C) - n(A\cap B\cap C) = 43 - 2 = 41n(A∪B∪C)−n(A∩B∩C)=43−2=41$

The number of people who like at most 2 is 41.