14. $S=\{1,2, \cdots 100\}, S$ has a 10-element subset called an alternating set if and only if, when the elements of the subset are arranged in ascending order, they form an alternating odd-even sequence. Find the number of all alternating subsets of $S$.
14. Let $A$ be the sub-cluster composed of all alternating sets of $S$, and $B$ be the sub-cluster of all 10-element subsets of $\{2,3, \cdots 110\}$ where all elements are either all odd or all even.
Define the mapping $f: A \rightarrow B$ as follows:
For any $\left\{a_{1}, a_{2}, \cdots a_{10}\right\} \in A\left(a_{1}<a_{2}<\cdots<a_{10}\right)$, then $\left.f\left(\mid a_{1}, a_{2}, \cdots, a_{10}\right\}\right)=$ $\left|b_{i}\right| b_{i}=a_{i}+i, i=1,2, \cdots, 10 \mid$

Since $a_{1}, a_{2}, \cdots a_{10}$ alternate in parity, $b_{1}, b_{2}, \cdots b_{10}$ have the same parity. It is easy to prove that $f$ is injective, and it is also easy to see that for any element in $B$, there exists a preimage in $A$ (since $110-10=100,2-1=1$), so $f$ is also surjective.
Therefore, $|A|=|B|$.
Since the elements of $B$ can only be 10-element subsets of $|2,4,6, \cdots 110|$ or $|3,5,7, \cdots 109|$, we have $|B|=C_{55}^{10}+C_{54}^{10}$.
Thus, the number of all alternating sets is $C_{55}^{10}+\mathrm{C}_{54}^{10}$.