I disagree with the other tutor's answer.

4 groups of 6 means you use only 24 of the 26 apples, and so you have 
to leave two apples behind.  

First let's count the number of ways in which the 4 groups are
ordered: first group, second group, third group, and fourth group.
Then we'll divide by 4! to get rid of the extra ones, since order
doesn't matter.

The ways to choose the two to leave behind are C(26,2)
For each of those ways to leave two behind,
there are C(24,6) ways to choose the first group, then
there are C(18,6) ways to choose the second group, then
there are C(12,6) ways to choose the third group, then
there are C(6,6) (or 1 way) to choose the 4th group.

That's C(26,2)C(24,6)C(18,6)C(12,6)C(6,6) ways if the
groups were in a certain order.  But since order doesn't
matter, ordering them causes each to be counted 4! times 
too many, so we divide by 4!:



3.12642348×1013

Edwin