Question
Download Solution PDFयदि एक समुच्चय A में 3 अवयव हैं और दूसरे समुच्चय B में 6 अवयव हैं, तो (A∪B) में न्यूनतम कितने अवयव हो सकते हैं?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFअवधारणा:
यदि A और B दो समुच्चय हैं तो, n (A ∪ B) = n (A) + n (B) - n (A ∩ B)
गणना:
दिया गया है: n (A) = 3 और n (B) = 6
जैसा कि हम जानते हैं कि, यदि A और B दो समुच्चय हैं तो, n (A ∪ B) = n (A) + n (B) - n (A ∩ B)
⇒ n (A ∪ B) = 3 + 6 - n (A ∩ B)
n (A ∪ B) को कम करने के लिए हमें n (A ∩ B) को अधिकतम करना होगा।
यदि A, B का उप-समुच्चय है, तो A ∩ B = A ⇒ n (A ∩ B) = n (A) = 3
⇒ n (A ∪ B) = 3 + 6 - 3 = 6
इसलिए, घटकों की न्यूनतम संख्या जो (A∪B) में हो सकती है, वह 6 है।
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