Question
Download Solution PDFक्रमशः 150, 180, 210 और 240 तत्व होनेवाले 4 समुच्चय A, B, C और D के संघ में तत्वों की संख्या को ज्ञात करें, जब दिया हुआ है कि समुच्चय के प्रत्येक जोड़े में 15 तत्व उभयनिष्ठ हैं। समुच्चय के प्रत्येक त्रिक (ट्रिपल) में 3 तत्व उभयनिष्ठ हैं और A ∩ B ∩ C ∩ D = ϕ।
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Detailed Solution
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दिया हुआ: चार समुच्चयों में क्रमशः 150, 180, 210 और 240 तत्व हैं
n(A) = 150
n(B) = 180
n(C) = 210
n(D) = 240
समुच्चय के प्रत्येक जोड़े में 15 तत्व हैं
n(A ∩ B) = 15
n(A ∩ C) = 15
n(A ∩ D) = 15
n(B ∩ C) = 15
n(B ∩ D) = 15
n(C ∩ D) = 15
समुच्चय के प्रत्येक त्रिक (ट्रिपल) में 3 तत्व हैं
n(A ∩ B ∩ C) = 3
n(A ∩ B ∩ D) = 3
n(A ∩ C ∩ D) = 3
n(B ∩ C ∩ D) = 3
A ∩ B ∩ C ∩ D = ϕ
n(A ∩ B ∩ C ∩ D) = 0
अब 4 समुच्चय A, B, C और D के संघ में तत्वों की संख्या
n(A ∪ B ∪ C ∪ D) = n(A) + n(B) + n(C) + n(D) - n(A ∩ B) - n(A ∩ C) - n(A ∩ D) - n(B ∩ C) - n(B ∩ D) - n(C ∩ D) + n(A ∩ B ∩ C) + n(A ∩ B ∩ D) + n(A ∩ C ∩ D) + n(B ∩ C ∩ D) - n(A ∩ B ∩ C ∩ D)
= 150 + 180 + 210 + 240 - 6 × 15 + 4 × 3 - 0
= 702
Last updated on Jun 12, 2025
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