Need help understanding probability proposition from Sheldon Ross I am reading Sheldon Ross's book 'A first course in Probability'. In the second chapter, he is describing multiple proposition. I cannot understand proposition 4.4. I have it pasted below:



I follow until the moment he introduces (m k)

Moved from Comment format to Answer format per suggestion.
Here is a guide to understanding the special case for
$P(A\cup B\cup C).$
Start with a Venn Diagram with three circles: one each for $A, B, C$ overlapping. I have in mind a diagram with $2^3 = 8$ regions, including the region outside all of the circles.
[Notice that there is ${3 \choose 0} = 1$ region containing
none of the three sets, there are ${3\choose 1} = 3$ regions containing parts of one set, ${3\choose 2} = 3$
(lens-shaped) regions containing parts of two sets, and there is ${3\choose 3} = 1$ (roughly triangular shaped) region containing parts of all three sets. Also, $1 + 3 + 3 + 1 = 2^3 = 8$ regions.]
If you add probabilities for $A$ and $B$ and $C$ to try for $P(A∪B∪C),$ you have double-counted three lens-shaped regions for $A∩B, B∩C$ and $A∩C$ and triple-counted one small almost triangular region for $A∩B∩C.$ So find $$P(A)+P(B)+P(C)−P(A∩B)−P(B∩C)−P(A∩C).$$
Now you have subtracted $P(A∩B∩C)$ three times. So you have $P(A∪B∪C)−P(A∩B∩C).$ Finally, to get $P(A∪B∪C)$ by adding back $P(A∩B∩C).$ In summary,
$$P(A\cup B\cup C) \\
= P(A)+P(B)+P(C)\;−P(AB)−P(BC)−P(AC)\;+P(ABC).$$
This process is called 'inclusion-exclusion'. To get $P(A∪B∪C)$ you have included three probabilities, excluded three smaller ones, and included one relatively very small probability.
The equation in Ross' book shows the inclusion-exclusion process necessary to get the general probability $P(E_1∪E_2,∪⋯∪E_n)$ with the proper number of inclusions, exclusions, inclusions, ..., and so on.
It helps to understand clearly what happens for three events before trying to generalize to $n.$
