Question 938036
The way I see it, choosing the {{{12+2=14}}} people who could form the final panel is the same as choosing the {{{20-14=6}}} people that are not wanted.
That is {{{20*19*18*17*16*15/(1*2*3*4*5*6)=19*17*15*8=38760}}} ,
so there are {{{38760}}} different sets of {{{14}}} people that can be picked.
For each of those {{{14}}} sets, there are
{{{14*13/(1*2)=7*13=91}}} possible sets of {{{2}}} alternates.
That gives us {{{38760*91=highlight(3527160)}}} possible panels that can be formed.
 
Another way:
Out of the {{{20}}} people available, there are
{{{20!/(8!12!)=125970}}} ways to chose the {{{12}}} jurors,
and in each case, from the {{{20-12=8}}} people remaining, there are
{{{8*7/(1*2)=4*7=28}}} ways to choose two alternates.
So, there are {{{125970*28=highlight(3527160)}}} possible panels that can be formed.