
Hi, I have a question to ask regarding subquestion 3. in the picture. I solved it by using ${10 \choose 2}$ since $2$ of the $4$ houses are fixed already, which I thought would leave me with $10$ choices. However, the answer is $126$, which completely puzzles me, because thinking back, I realised that the $10$ houses are of $5$ different designs, which made me wonder if my answer was wrong. Even so, my method would have overcounted, but the answer is way more than mine, which means I have undercounted instead. I simply cannot find out why am I wrong.