Here is some psuedo-code
function player_combos(array players, int minimum_total) {
array result = []
players = sort(players, metric=most points first)
int total = 0
for p1 in 0 .. length(players) {
if players[p1].points*10 < minimum_total - total:
break
total += players[p1].points
for p2 in p1+1 .. length(players) {
if players[p2].points*9 < minimum_total - total:
break
total += players[p2].points
for p3 in p2+1 .. length(players) {
if players[p3].points*8 < minimum_total - total:
break
total += players[p3].points
# continue these nested loops up to p10
...
for p10 in p9+1 .. length(players):
if players[p10].points < mininum_total - total:
break
# this is a valid combination
result.append((p1, p2, p3, p4, p5, p6, p7, p8, p9, p10))
...
# remember to decrement total when we finish a loop iteration
total -= players[p3].points
}
total -= players[p2].points
}
total -= players[p1].points
}
return result
}
The idea here is that because you have the players sorted first, at any point while looping over players in the list, all players after must have an equal or lower point total as the current player. This allows you to break out of the current loop if the current player's points multiplied by the number of remaining spots left on the team is less than the number of points required to meet the minimum.
For example say you have four players on the team so far with 80 points total, that means you have 90 points left to reach the minimum, and there are 6 spots left. The absolute minimum number of points that your next player can have is 15 (since 90 / 6 == 15), so as soon as you reach a player in next loop that has 14 or fewer points, you can break out of that loop.
This should drastically reduce the total number of combinations you need to get, as long as your minimum_total metric is set high enough.
