### Video Transcript

The diagram shows a logic circuit
consisting of three OR gates. How many different possible
combinations of input values are there for this circuit?

We can see that we’ve got a diagram
here that shows a logic circuit which contains three OR gates. The circuit has four inputs, and
these are labeled as 𝐴, 𝐵, 𝐶, and 𝐷. The inputs 𝐴 and 𝐵 are the two
inputs to this upper-left-hand OR gate, while inputs 𝐶 and 𝐷 go into the
lower-left-hand OR gate. The outputs from these two gates on
the left then become the two inputs for this third OR gate on the right-hand side of
the circuit. Finally, the output from this
right-hand OR gate is the overall output of the logic circuit.

As it turns out in this particular
question, we’re not actually concerned about this output value at all. Instead, the question is just about
these four inputs. And specifically, we’re asked to
work out how many different possible combinations of input values that there
are. To answer this question, we need to
recall that whenever we’re talking about a logic circuit, there are only two
possible values, and these are zero and one. So, this means that each of the
inputs to this circuit can either have a value of zero or a value of one.

So, input 𝐴 can have a value of
zero or a value of one. And completely independent from the
value of 𝐴, input 𝐵 can also be either zero or one. And again, independent of the
values of the other inputs, input 𝐶 can be either zero or one and input 𝐷 can
separately be either zero or one. So then, we’ve got four different
inputs, and there are two possibilities for the value taken by each of these
inputs.

We want to work out the number of
different possible combinations of these four input values that there are. Now, there are a couple of
different ways that we could go about this. One approach would be to
methodically go through and explicitly write out all of the different combinations
and then count them up. However, not only is this approach
quite a time-consuming one, but there’s also the possibility of accidentally missing
out one of the different combinations. And if we did this, we would get
the wrong answer.

Luckily, there’s a better approach
that we can use. We know that we have four different
inputs, each of which can take either of two possible values. These inputs are all totally
independent, which means, for example, that whether input 𝐴 is equal to zero or to
one doesn’t affect the potential values for inputs 𝐵, 𝐶, and 𝐷. These can each independently be
equal to either zero or one. Then, the total number of different
possible combinations must be equal to the two possibilities for input 𝐴 multiplied
by the two possibilities for input 𝐵 multiplied by the two possibilities for input
𝐶 and finally multiplied by the two possibilities for input 𝐷. So, we’ve got that this total
number of combinations is equal to two times two times two times two. That’s one factor of two for each
of the four inputs to the logic circuit.

Now, we could simply go ahead and
evaluate this expression. However, we can first gain a little
bit more insight into the problem by noticing that we can rewrite this product of
four factors of two as two raised to the power of four. And now, if we look at this
expression for the number of different combinations, we can identify two as the
number of different values that each input could have because each of the inputs
could either be zero or one. We can also identify four, the
power that two is raised to as the number of inputs because we’ve got the four
inputs 𝐴, 𝐵, 𝐶, and 𝐷.

In fact, as a bit of an aside, this
result that we found can be generalized. If instead of four inputs, we had a
logic circuit with some general number of inputs, which we’ve labeled as 𝑁, then
the number of different possible combinations of the values of these inputs would be
equal to two raised to the power of 𝑁. That’s because each of the 𝑁
inputs could have either of the two values zero or one. So then, just as we found for four
inputs the number of combinations was equal to two times two times two times two,
that’s four factors of two and this gave us two to the power of four, then for 𝑁
inputs, we’d end up with a product of 𝑁 factors of two, which we could write as two
to the power of 𝑁.

Now, getting back to the question,
in our case of four inputs, we know that the number of combinations is equal to two
raised to the power of four. Evaluating this gives us our answer
that the number of different possible combinations of input values for this circuit
is equal to 16.