# Pre-RMO 2014/20

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 $B \subseteq A$ ?

This note is part of the set Pre-RMO 2014

Note by Pranshu Gaba
4 years, 8 months ago

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## Comments

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is it 30???

- 4 years, 8 months ago

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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

- 4 years, 8 months ago

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

- 4 years, 8 months ago

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

- 4 years, 8 months ago

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should be 1024 - 243 - 243 + 32 = 570

- 4 years, 8 months ago

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

- 4 years, 8 months ago

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

- 4 years, 8 months ago

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

- 4 years, 8 months ago

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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 $x$.

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

- 4 years, 8 months ago

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

- 10 months, 3 weeks ago

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

- 4 years, 8 months ago

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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 = $2^5 \times 2^5 = 2^{10}$

- 4 years, 8 months ago

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6

- 4 years, 8 months ago

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570

- 3 years, 8 months ago

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

- 3 years, 8 months ago

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