The number of ways of seating three gentleman and three ladies in a row ,such that each gentleman is adjacent to at least one lady The number of ways of seating three gentleman and three ladies in a row ,such that each gentleman is adjacent to at least one lady, is **_My try :_** If we try to find the complement of the situation and subtract it from 6! then it may be easier. If all the three gentleman are together then at least one will not be adjacent to any lady. If all the three ladies are between any two gentlemen then also the other gentleman is not adjacent to any lady. But my counting showing error. Please do provide any handy method else let me know if I missed any other cases. Help appreciated :)

See we will go by your method using complements. See total ways are $720$ . now lets group three men so let us assume them $x$ . now let us place them in positions $123$ so total ways are $3!.3!=36$ as men and women can e arranged amongst in $3!$ ways. Now place $x$ at place $234$ again same number of ways ie $36$ now similarly at $345,456$ so total ways become $36\cdot 4=144$ now group any two men . this can be done in $3$ ways. Place them at place $1,2$ now total ways become $3.2!.3.3!=108$ . men can be arranged in $2!$ ways $3rd$ place has to be female or it becomez similar to $3$ men together so $3$ ways and $1M,2w$ can be arranged in $3!$ ways. Note that now they cant be placed in positions $(23),(34),(45)$ as it becomes similar to $3$ men together. So now they can be placed at $56$ position. Again similar to $12$ position we get $108$ ways . now remember we are talking complement of total permitted ways . hence total ways are $720-144-216=360$