Number of ways in which 7 green bottles and 8 blue bottles can be arranged in a row if exactly 1 pair of green bottles is side by side, is (Assume all bottles to be alike except for the colour):

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$$\begin{gathered} \text{ 8 identical blue bottles can be arranged in one way} \\ \mathrm{There} \mathrm{are} 9 \mathrm{gaps} \mathrm{before} \mathrm{between} \mathrm{and} \mathrm{after} \mathrm{blue} \mathrm{bottles} \\ {\text{One gap out of nine can be taken in }}^9{\mathrm{C}}_1\text{ ways and }2\text{ green bottles can be placed. } \end{gathered}$$

Now,   for remaining 5 green bottles there are 8 gaps$5\text{ gaps out of }8\text{ gaps for remaining }5{\text{ green bottles can be selected in }}^8{C}_5\text{ . }$

$\begin{array}{l}\text{ Hence, total ways }{=}^9{C_1}^8{C}_5=9\cdot \frac{\mathrm{8.7.6}}{\mathrm{1.2.3}}\\ =72.7=504\end{array}$