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If n(r' intersection s') + n(r' intersection s)=3, n(r intersection s)=4 and n(s' intersection r)=7​

User KoreanDude
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We can use the principle of inclusion-exclusion to find n(U), which states that for two sets A and B:

n(A union B) = n(A) + n(B) - n(A intersection B)

We can apply this to three sets r, s, and their complements r' and s':

n(U) = n(r union s)

= n(r) + n(s) - n(r intersection s)

= [n(r intersection s') + n(r intersection s)] + [n(s intersection r') + n(s intersection r)] - n(r intersection s)

= [(4 + n(r' intersection s)) + (n(r intersection s') + 7)] - 4

= n(r' intersection s) + n(r intersection s') + 3

= 7 + 3 + 3

= 13

Therefore, n(U) = 13.

User Nerdybeardo
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