Question 815307
There are 66 ways to chose 2 out of  a group of 12 if the order in which they are chosen does not matter:{{{12*11/2=66}}} .
 
There are 120 ways to chose 3 from the remaining 10,
{{{12-2=10}}} ,
if the order in which they are chosen does not matter:
{{{10*9*8/3/2=120}}} .
 
For each of the calculations above the math jargon is combinations of m elements taken from a set of n elements, calculated as {{{n!/(n-m)!m!}}}={{{ n*(n-1)(n-2)*"...."*(n-m+2)*(n-m+1)/(m*(m-1)(m-2)*"...."2*1))}}}
 
So, there are 7920 ways to take a group of 2 and a group of 3 out of 12, leaving a group of 7 remaining:
{{{66*120=highlight(7920)}}}  .