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If the set contains one number, then it must be {5}.
If the set contains three numbers, then it must contain 5, one number less than 5, and one number greater than 5. There are 4 numbers less than 5 and 6 numbers greater than 5 to choose from, giving
a total of 4×6 = 24 sets.
More generally, if the set contains \(2n+1\) numbers, then it must contain 5, \(n\) numbers less than 5, and \(n\) numbers greater than 5.
There are \(\left(\begin{array}{c}4\\n\end{array}\right)\) numbers less than 5 and \(\left(\begin{array}{c}6\\n\end{array}\right)\) numbers greater than 5 to choose from, giving
a total of \(\left(\begin{array}{c}4\\n\end{array}\right)\times\left(\begin{array}{c}6\\n\end{array}\right)\) sets.
In total there are
$$
1+24+
\left(\begin{array}{c}4\\2\end{array}\right)\times\left(\begin{array}{c}6\\2\end{array}\right)
+
\left(\begin{array}{c}4\\3\end{array}\right)\times\left(\begin{array}{c}6\\3\end{array}\right)
+
\left(\begin{array}{c}4\\4\end{array}\right)\times\left(\begin{array}{c}6\\4\end{array}\right)
$$
sets. This is equal to 210.
