Why is the resistance in a parallel circuit smaller than the resistance in a series circuit? So I was told in the physics class that the resistance in a parallel circuit is smaller than the resistance in a series circuit. Why does that happen?
Is this statement also true for circuits which have no resistors or resistance-offering devices connected to them?
And also from my calculations the total resistance in a parallel circuit is smaller than the resistances of any of the devices connected in the circuit. Now I really don't understand how this can happen. 
 A: Suppose you have a voltage $V$ between two points A and B in a circuit.  If initially you have a resistor of resistance $R_1$ between A and B, the current flowing through the resistor is $I_1=V/R_1$.  Now if you connect another resistor $R_2$ in parallel to the resistor $R_1$, then the former will have the same voltage $V$ across it (since it is connected to the same points A and B).  Therefore the current through $R_2$ will be $I_2=V/R_2$.  Therefore, the total current $I_{tot}$ entering at A and leaving at B will be
$$I_{\text{tot}}=I_1+I_2$$
$$I_{\text{tot}}=V\left(\frac{1}{R_1}+\frac{1}{R_2}\right) $$
$$I_{\text{tot}}=\frac{V}{R_{\text{eff}}}$$
where $$ \frac{1}{R_{\text{eff}}}=\frac{1}{R_1}+\frac{1}{R_2} $$
$R_{\text{eff}}$ is the "effective resistance" between the points A and B.  It determines what current passes through the points A and B.  You can extend this concept to more resistances added in parallel between A and B.
The value of $R_{\text{eff}}$ will be smaller that the value of the smallest resistance between A and B.  You can deduce this as follows:
If $$R_1<R_2$$
$$ \frac{1}{R_1}>\frac{1}{R_2}$$
$$\frac{1}{R_1}+\frac{1}{R_2}>\frac{1}{R_1}$$
$$\Rightarrow \:\: R_{\text{eff}}<R_1$$
When you are considering DC, it does not really make sense to calculate the effective resistance when you have something other than a resistor which does not have resistance, connected in parallel.  For example, an inductor connected in parallel will just cause a short circuit whereas a capacitor will cause an open circuit.  
But when considering AC, things are a lot more interesting and different.  The concept of complex impedances is then used. See http://hyperphysics.phy-astr.gsu.edu/hbase/electric/impcom.html for more details.
A: If I have a circuit with resistance R and voltage V, I get a certain current - that's Ohm's law, $I = \frac{V}{R}$.
Now imagine you have two such circuits - completely separate from each other. Each will have the same current. Let's say the voltage is 1 V, and the resistance is 1 A:

Now if I connect the terminals of the two voltage sources together (which I can do because they have the same voltage) I get this:

Because the voltage sources are the same, I can remove on to get this:

And now I have two resistors in parallel, carrying twice as much current as a single resistor. From the perspective of the entire circuit, the "effective resistance" is half (because, for the same voltage, the current is double).
Now let's put the resistors in series. I use the same trick: initially I have separate circuits, and then I connect them:

The voltage drop across each resistor is the same; there is an equal current flowing "left" and "right" in the wire in the middle, which I can therefore remove (it carries no net current). I am left with a circuit that carries one amp, but it has two resistors in series, and two voltage sources. In other words - in order to maintain the current in a series resistance network, I have to double the voltage when I put two resistors in series.
The rest (getting the equations for the general case of any number of resistors or unequal size) is just math...
A: Let the potential difference $V$ is equal to sum of potential difference $V_1,V_2...$
In series connection
$$V=V_1+V_2..$$
$$V=IR_1+IR_2...$$
$$IR=IR_1+IR_2...$$
Therefore the equivalence resistance,
$$R_S=R_1+R_2...$$
,which always greater than individual resistance
While in parallel combination
$$I=I_1+I_2+...$$
$$=>~~~~I=\frac {V}{R_1} +\frac {V}{R_2} + ...$$
Since $I=\frac{V}{R_p}$
$$\frac{V}{R_p}=\frac {V}{R_1} +\frac {V}{R_2} + ...$$
The the total resistance in  a parallel combination,
$$\frac {1}{R_{p}}=\frac {1}{R_1} +\frac {1}{R_2} + ...$$
,which always less than individual resistance .
Therefore obviously less than series combination.
A: It seems like you get the rational answer but lack the ability to feel it:
imagine you have 5 doors of different sizes and a thousand people to pass from a to b. If all the doors are in a row (so everyone has to pass every door), it will take a lot more time compared to the situation, where you place all the doors next to each other so that every person can choose to go either through door 1 or 2 or 3... which is called parallel.
Disclaimer: this example lacks the ability to count for the adding resistance in a serial circuit and is not a full answer to your question but a way of getting an idea, why this somehow paradox (I know, it feels somehow like that) behavior occurs.
A: From the wording of the question I would assume the OP didn't want formulas or a very technical answer, so I'll attempt answering in layman's terms.
What does resistance do? It resists the flowing of current. Given the same voltage, the bigger the resistance, the smaller amount of current can flow.
Now, imagine that there is a resistor. You put another one in parallel. Now, current can flow through both ways, so it's easier for current to flow. Imagine that previously there was only one lane on the highway, now there are two. More cars can get through during the same time. So, parallel resistors allow more current. This means that the system as a whole can let more current through, so its resistance is lower.
Now, imagine two resistors in series, after each other. Current has to pass through both of them, and both will resist the flow of current. So less current can pass through, therefore the system as a whole "resits the current more".
A: Question: Which of two pipes of equal length offers less resistance to the flow of water, one of which has twice the cross sectional area of the other?
Answer: The one of twice the cross sectional area.
But the one of twice the cross sectional area can be thought of as two of the smaller cross sectional area pipes in parallel.
This analogy gives an idea of the smaller resistance of a parallel arrangement of resistors although not completely.
A length $l$ of wire of cross sectional area $A$ has a resistance $R$ given by the equation $R = \dfrac {\rho l}{A}$ where $\rho$ is the resitivity of the wire.
Doubling the cross sectional area of a fixed length of wire decreases the resistance by a factor of two which is equivalent to having two wires of area $A$ in parallel.
However for a pipe the "resistance" to fluid flow is proportional to $\dfrac {1}{\text{area}^2}$.
So in the case of a pipe having the area increase by a factor of two decreases the resistance to fluid flow by a factor of four.
A: When you ask:

So I was told in the physics class that the resistance in a parallel circuit is lesser than the resistance in a series circuit.

This question only applies when two resistors are connected in series, versus the same two resistors connected in parallel instead.  It's important to understand that apples-to-apples comparisons can only be made when the circuits differ only in how the resistors are connected.
Now, given this, your next statement:

Why does that happen?

Convincing someones else "why" something is true in physics is a really tough job for anyone.  However, I recommend sitting down with a calculator and calculating the total resistance of two $5 \Omega$ resistors in parallel, and then comparing that with the total resistance of two $5 \Omega$ resistors in series.  Then repeat the calculations with $4 \Omega$ resistors, $10 \Omega$ resistors, and so forth until you convince yourself (as I originally did) that the statement must be true for all resistors.
You go on to ask the question:

Is this statement also true for circuits which have no resistors or resistance offering devices connected to them?

Circuits that have no resistance at all (on paper) have an infinite amount of current going through them (on paper).  Trying to compare this situation with circuits that have resistance is a fairly meaningless comparison.  If this is a homework question, I recommend answering "no" and then stating that these two situations cannot be meaningfully compared.
Lastly, you go on to ask:

And also from my calculations the total resistance in a parallel circuit is lesser than the resistances of any of the devices connected in the circuit.

Again, it might resolve the confusion is you simply tried out various resistor situations, did the calculations with a calculator until you realize that it must be true for all resistor values.
