SOLUTION: Three ladies have each brought a child for admission. The head of the school wishes to interview the six people in such a manner that no child is interviewed before his/her mother.

A mother must interview first.
Then either her child or another mother must interview second. 

Case 1. Her child interviews second 
     Then another mother must interview 3rd.
     Then the last 3 interviewees can occur 3 ways, 
     CMC which can occur only one way, or
     MCC which can occur 2 ways.

That's 3 ways for case 1.

Case 2: Two mothers interview first and second.

     Sub-case 2a: A mother interviews third.
     The 3 children can interview in any of 3! or 6 ways. 
     That's 6 ways for sub-case 2a.

     Sub-case 2b: A child interviews third.
     We can choose that child 2 ways, as the first or second 
     interviewing mother's child.
     Then for each of those 2 ways, the last 3 interviewees 
     can occur 3 ways.  That is, 
     CMC which can occur only one way, and
     MCC which can occur 2 ways.
     So there are 2*3=6 ways for sub-case 2b

So there are 6+6 = 12 ways for case 2.

That's a total of 3+12=15 ways for the scheme of interviewing.

Then there are 3!=6 ways the three mother-child pairs can be 
permuted.

Final answer: 15*6 = 90 ways.

Edwin