您可以尝试此查询：

match (u:User)-[:SUBSCRIBE]->(c:Channel) WITH u, collect(id(c)) AS userChannels, collect(c.name) AS channelNames
match (c:Channel)
with u, userChannels, channelNames, collect(id(c)) as channels
with apoc.coll.combinations(channels, 1, size(channels)) as combinations, u, userChannels, channelNames
unwind combinations as combination
WITH u, userChannels, channelNames, combination WHERE size(combination) = size(userChannels) AND ALL(x IN userChannels WHERE x IN combination)
RETURN channelNames, count(u) AS usersCount

在这个查询中，我们在两个列表中为每个用户收集频道ID和名称。然后，我们计算所有的通道节点组合。在解开组合后的最后，我们检查combination列表和userChannels列表的大小是否相同，如果是，我们比较内容。最后，对于每个组合，我们返回用户数。

要回答neo4j是否是适合这项工作的工具，可以提供数据集在某种意义上是有界的。当前的查询基本上是遍历数据库中存在的每个节点和关系，而且随着时间的推移，组合也会增加，因此扩展可能会成为一个问题。

尝试此查询，这将给出预期的计数：

match (c:Channel)
with collect(c.name) as channels
with apoc.coll.combinations(channels, 1, size(channels)) as combinations
match (u:User)-[:SUBSCRIBE]->(c:Channel)
with u, collect(c.name) AS userChannels, combinations
UNWIND combinations AS combination
WITH combination, u WHERE size(combination) <= size(userChannels) and apoc.coll.containsAll(userChannels, combination)
return combination, count(distinct u)