Access的count distinct的方法

网上查了很多关于ACCESS的COUNT DISTINCT的方法，都不好使，在一个老外的贴子上找到了方法，贴出来
SELECT C.immu, T2.N
FROM [SELECT T.immu, Count(T.immu) AS N
     FROM
        (SELECT DISTINCT immu, Count(*) AS N
         FROM plan GROUP BY immu) AS T
    GROUP BY T.immu]. AS T2 INNER JOIN plan AS C ON T2.immu = C.immu
GROUP BY C.immu, T2.N;