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# Understand the problem

True or False? Let $$H_1, H_2, H_3,H_4$$ be four hyperplanes in $$R^3$$. The maximum possible number of connected components of $$R^3 – H_1 \cup H_2 \cup H_3 \cap H_4$$ is 14.
##### Source of the problem
TIFR GS 2017 Entrance Examination Paper
General Topology
Easy
##### Suggested Book
Topology, Second Edition, English, Paperback, by James R. Munkres

Do you really need a hint? Try it first!

If $$m$$ is the number of hyperplanes and $$n$$ is the dimension of the space in which the hyperplanes are intersecting then can you derive a formula which will give you the maximum possible number of connected components?
When $$m=2$$ and $$n=1$$ (that is, two hyperplanes are passing through a line, say) then the number of maximum possible connected components is $$2$$, that is $${m \choose n}$$.
When $$m=3$$ and $$n=2$$ (that is, three hyperplanes are intersecting in $$\Bbb R^2$$ space) then the number of maximum possible connected components is $$3$$, that is $${ m \choose n}$$.
Hence, the maximum possible number of connected components when $$m$$ hyperplanes are intersecting in $$\Bbb R^n$$ is $${m \choose 0}+{m \choose 1}+{m \choose 2}+{m \choose 3}+ \ldots +{m \choose n}$$ when $$m>n$$ and is $$2^m$$ when $$m<n$$ and $$2^m-1$$ when $$m=n+1$$.
Now, what do you think could be the maximum possible number of connected components in the given question?
The maximum possible number of connected components is $$15$$. But the statement is saying $$14$$. Hence the statement is false..

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