【Example 6】There are three sets of real numbers: (A) $a_{1}<a_{2}<\cdots<a_{n_{1}}$; (B) $b_{1}<$ $b_{2}<\cdots<b_{n_{2}}$; (C) $c_{1}<c_{2}<\cdots<c_{n_{3}}$. Now we want to interweave them into one sequence, but the increasing order of each set must be maintained. How many different sequences can be formed?
Since the three sets of real numbers must all remain in increasing order, there is no need to consider the internal arrangement order of each set, but only to consider which positions the three sets are respectively listed. Therefore, it is still equivalent to a permutation problem of not entirely different elements. Hence, we know that there can be
$$
\left(\begin{array}{ccc}
n_{1}+n_{2}+n_{3} \\
n_{1} & n_{2} & n_{3}
\end{array}\right)=\frac{\left(n_{1}+n_{2}+n_{3}\right)!}{n_{1}!n_{2}!n_{3}!}
$$

different sequences.