What is the number of ordered pairs \((A, B)\) where \(A\) and \(B\) are subsets of \(\{1, 2, \ldots, 5\}\) such that neither \(A \subseteq B\) nor \(B \subseteq A\) ?

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## Comments

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TopNewest@Pranshu Gaba : Are you studying in resonance? Which Centre?

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570

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6

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why should it be 2^10......i think it should be ${2^5 \choose 2}*2$

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Using rule of product, \(A\) has \(2^5\) choices, and even \(B\) has \(2^5\) choices, so total no. of ordered pairs = \(2^5 \times 2^5 = 2^{10}\)

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As @ww margera said, we can use principle of inclusion and exclusion.

\[ \text{Answer} = \text{total no. of ordered pairs} - (|A \subseteq B |+ |B \subseteq A|) + |A = B| \]

where \(|x|\) is no. of ordered pairs \((A, B)\) satifying condition \(x\).

\[ \text{Answer}= 2^{10} - ( 3^5 + 3^5) + 2^5 = \boxed{570}\]

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How 2^10, 3^5, 2^5, I didn't get that.

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@ww margera. actually you permuted it...... {1,2} is Same as {2,1}

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should be 1024 - 243 - 243 + 32 = 570

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ww margera. actually you permuted it...... {1,2} is Same as {2,1}

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Ordered pairs right? So the two should not be the same...

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Write a comment or ask a question... I think 52

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How many elements can A or B can have? Only 1 or more than one

If only 1 element ans \(\to\) 20

Otherwise(counting every possibility) ans \(\to\) 47

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Since \(A\) and \(B\) are subsets of \(\{1, 2, 3, 4, 5\}\), they can have from 0 to 5 elements.

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is it 30???

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