\[ \text{Make an ordered selection to place 4 out of 10 in row 1: }^{10}P_4.\\ \text{Then make an ordered selection to place 4 out of 6 in row 2: }^{6}P_4\\ \text{And finally make an ordered selection to place 2 out of 2 in row 3: }^{2}P_2 \]
\[ \text{However we haven't been told that the rows are distinguishable.}\\ \text{Therefore we would've treated arrangements of the form}\\ \boxed{ABCD}\, \boxed{EFGH}\, \boxed{IJ}\\ \text{identical to the arrangement}\\ \boxed{EFGH}\,\boxed{ABCD}\,\boxed{IJ}.\]
\[ \text{Hence to address the double counting for the two identical rows}\\ \text{an appropriate correction factor of }\frac{1}{2!}\text{ is required.}\\ \text{This gives an answer of }\frac{1}{2!} \, ^{10}P_4\, ^{6}P_4 \, ^2P_2\]
This should be identical to \( \frac{10!}{2!} \) I think