Two resistors can be wired two ways, and the difference is not cosmetic. In series they sit end to end, so the same current must cross both in turn. In parallel they sit side by side between the same two nodes, so the same voltage appears across both and the current chooses.

Series first: $R_t = R_1 + R_2$, read aloud R-total equals R-one plus R-two. $R_t$ is the total resistance the rest of the circuit sees, in ohms; $R_1$ and $R_2$ are the two components, also in ohms, and the subscripts are just names — swapping them changes nothing. Every ohm in the path is an ohm the current has to get through, so the total is bigger than either one.

Parallel: $R_t = \dfrac{R_1 R_2}{R_1 + R_2}$ — R-total equals R-one R-two over R-one plus R-two, the product-over-sum shortcut, which is the reciprocal rule $\dfrac{1}{R_t} = \dfrac{1}{R_1} + \dfrac{1}{R_2}$ already solved for two branches. Use the shortcut for two and the reciprocal form for three or more.

Here is the sanity rail, and it is worth more than the arithmetic: a parallel combination always lands BELOW the smaller of the two branches. Not between them, not near their average — below the smaller. Opening a second road can only make the journey easier. Check every parallel answer against that rail before you write it down, and a swapped rule can never survive to the bottom of the page.