P, Q and R are three sets and ξP∪Q∪R.Given that n(ξ) = 60, n(P∩Q)=5, n(Q ∩R) = 10, n(p) = 20 and n(Q) = 23,find n (P' ∪ Q)

n(Q)−n(P∩Q)−n(Q∩R)=23−5−10=88+ n(P)+n(R)=60n(R)=60−8−20=32n(P′∪Q)=n(R)+n(Q)−n(Q∩R)=32+23−10=45

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P, Q and R are three sets and P Q R. Given that n() = 60, n

Question

P, Q and R are three sets and $ξ P \cup Q \cup R .$ Given that n $(ξ)$ = 60, $n (P \cap Q) = 5$ , n(Q $\cap R$ ) = 10, n(p) = 20 $an d$ n(Q) = 23, $f in d$ n (P' $\cup Q$ )

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Sebastian Discussed

P, Q and R are three sets and

ξ P \cup Q \cup R .

Given that n

(ξ)

= 60,

n (P \cap Q) = 5

, n(Q

\cap R

) = 10, n(p) = 20

an d

n(Q) = 23,

f in d

n (P'

\cup Q

)

5 mins ago

Discuss this question LIVE

5 mins ago

n (Q) - n (P \cap Q) - n (Q \cap R) = 23 - 5 - 10 = 8

8 + n (P) + n (R) = 60

n (R) = 60 - 8 - 20 = 32

n (P^{'} \cup Q) = n (R) + n (Q) - n (Q \cap R)

= 32 + 23 - 10 = 45

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Question Text	P, Q and R are three sets and $ξ P \cup Q \cup R .$ Given that n $(ξ)$ = 60, $n (P \cap Q) = 5$ , n(Q $\cap R$ ) = 10, n(p) = 20 $an d$ n(Q) = 23, $f in d$ n (P' $\cup Q$ )
Answer Type	Text solution:1
Upvotes	150