Binomial expansion through combinations. If you have $(a+b)(c+d)(e+f)$ how can you expand this? Someone was mentioning that you get different combinations so that you get $adf+ade+acf+ace+bdf+bde+bce+bcf$? Is that the full expansion?
As an example, how can it be applied to the boolean expression:
$F' = [x'w'+z]*[x'+y'z']*[x'+z']$ which should reduce to the sum of products?
 A: You will indeed get $2^3 = 8$ terms when expanding a triple binomial: so yes, the first expansion is indeed the full expansion, barring any information about the relation of the variables $a, b, c, d, e, f$.
It can be applied to your boolean expression $F'$. Based on the given "standard expansion, we obtan:
$$\begin{align} F' & = x'w'y'z' + x'w'z' +  x'w'x' + x'w'x' + zy'z' + zz' + zx' + zx'\\ \\
& = x'w'y'z' + zx'\tag{1}\end{align}$$
Can you see how we can reduce the original expansion to two terms, as in $(1),\,$ from the $\,8$-term "formulaic" expansion? Here we have repetition of some of the variables, and we have the appearance of the variables $x, x'$ and $z, z'$:


*

*Note that there is reiteration (and hence we can simplify): $x'w'x' + x'w'x' = x'w'x' = x'w'$, and if the first term $x'w'y'z'$ holds, then so does $x'w'z'$ and so does $x'w'$. Similarly, there is reiteration/simplification: $zx' + zx' = zx'$

*The term $zy'z' = zz'y = F\cdot y = F$, and obviously, $zz' = F$.
So the reduced Sum of Products (which is essentially, disjunctive normal form) of $F'$ is expressed in $(1)$: $$F' = x'w'y'z' + zx'$$
