Question 1157980
I believe the answer expected would be {{{120!/12!^2/10!}}}
If you start by picking a group of 12 people for one of the committees, you have {{{120*119*118*"..."*110*109}}} ways to make a list,
but there could be other lists with the same names in different orders, for a total of
{{{12*11*10*9*8*7*6*5*4*3*2=12!}}} such lists,
So there are {{{120*119*118*"..."*110*109/12!}}} ways to pick the first group of 12 from the 120 organization members.
You could then pick from the 108 remaining members a second group of 12 in
{{{108*107*106*"..."*98*97/12!}}} ways,
then you could pick from the 96 remaining members a group of 12 in
{{{108*107*106*"..."*98*97/12!}}} ways,
then you could pick from the 96 remaining members a group of 12 in
{{{96*95*94*"..."*86*85/12!}}} ways,
then you could pick from the 84 remaining members a group of 12 in
{{{84*83*82*"..."*74*73/12!}}} ways,
And so on until you have used all 120 members of the organization and you have a list with
{{{120!/12!^2}}} lists of 10 groups in the order picked.
If the first entry in your is
(group A, group B, group C, group D, group E, group F, group G, group H, group I, group J) picked in that order,
the same set of groups appears in {{{10!}}} different orders down your list. Each set of 10 groups appears {{{10!}}} times, just with the group listed in different orders,
and nowhere it says that the order the committees were chosen matters, so we have to divide {{{120!/12!^2}}} by {{{10!}}} .