I am Vladimir Podolskii and today we are going to discuss Double counting.

Let's start with the following general idea.

It is useful to look at the same problem from two different angles.

Indeed, this way we can get information

from two different sources and sometimes it is crucial.

There is a special case of this simple idea in mathematics to

look at some number at your problem from two different points of view.

Let's consider the following example.

Is it possible to fill a table three by five by integers in such a way

that some of each integers in each row is equal to 20

and the sum of integers in each column is equal to 10?

Let's try to do it. So, lets fill the first row in some way.

So in the end we have to be careful,

so that the sum of the first row is 20.

Now, let's fill the second row in same way again and then we have to be careful

so that the sum is equal to 20.

Now, we are going to fill the third row,

and now we have to be careful every time because in each column we have to have 10,

so here we have to put three,

then three again, then four,

then zero and then we have to put zero since the sum of the columns should be 10.

On the other hand,

we see that the sum of the row is 10 and so we failed.

Let's try to show if this is impossible and how we are going to do it.

We assume that this is possible,

if we can fill the table with integers so we use proof by contradiction here.

Let's look the crucial idea here.

Let's look at the sum of all numbers in this table.

From one hand, the sum of each row is 20 so in total we have three rows.

So in total we have the sum of all numbers in this table is equal to 60.

On the other hand,

the sum of each column is 10,

so the sum of all numbers in all columns is equal to 50.

On one hand, the sum of all numbers in the table is 60,

and on the other hand it is 50.

So, this is a contradiction.

We have shown that this is impossible and let's summarize with how we have done this.

We have shown impossibility with calculating the same value

in two different ways and obtaining two different answers.

So, the key difficulty here or the main problem is to find your right value to compute.