I seem to have come rather late to this long discussion. Here's how I count the possible cases.

Case 1. Six different people receive an apple. There are $\displaystyle \textstyle {10\choose6} = 210$ ways of choosing which six. The other four must receive an orange, and there are then two spare oranges to allocate. There are $\displaystyle \textstyle {10\choose2}= 45$ ways of allocating them to two different people, and $\displaystyle \textstyle {10\choose1} = 10$ ways of giving them both to the same person. Total: $\displaystyle 210(45+10) = 11\,550$ ways.

Case 2. Five different people receive an apple. There are $\displaystyle \textstyle {10\choose5} =252$ ways of choosing which five. The other five must receive an orange. The remaining apple must go to one of the five who already have an apple (otherwise there would in fact be six different people with an apple, and we would be back in Case 1), so there are 5 ways of allocating it. The remaining orange can go to anyone, so there are 10 ways of allocating it. Total: $\displaystyle 252\times5\times10 = 12\,600$ ways.

Case 3. Four different people receive an apple. There are $\displaystyle \textstyle {10\choose4} = 210$ ways of choosing which four. The other six must each receive an orange. The remaining two apples must be allocated among the four who already have an apple. There are $\displaystyle \textstyle {4\choose2} = 6$ ways of giving an extra apple to two of them, and 4 ways of giving both apples to the same person. Total: $\displaystyle 210(6+4) = 2\,100$ ways.

The sum of the three cases gives a grand total of 26 250 ways, in agreement with previous answers above.