Permutations/Combinations...

well...the club is not registered.. it is surely different from this one.

mrblack ..could you please elaborate. I am getting your point. but can you solve the problem with the steps?

This is a complex question and I don't think the answer choices are correct.

The problem is that in a game, a person cannot be on two teams at the same time.

the number of 2 person teams from a group of 8 people is 8C2 = 28

If you want to visualize it:
let 8 people be represented by: a b c d e f g h

the teams they can form:
ab ac ad ae af ag ah
bc bd be bf bg bh
cd ce cf cg ch
de df dg dh
ef eg eh
fg fh
gh

28 teams

Each team can combine with one other team to form a game. (a person can only be on one team at a time):

ab can only play with teams that don't have a or b:
abcd abce abcf abcg abch abde abdf abdg abdh abef abeg abeh abfg abfh abgh

ac can only play with teams that don't have a or c:
acbd acbe acbf acbg acbh acde acdf acdg acdh acef aceg aceh acfg acfh acgh

ad can only play with teams that don't have a or d:
adbc adbe adbf adbg adbh adce adcf adcg adch adef adeg adeh adfg adfh adgh

and so on. so for each team in the first row, they can play with 15 other teams ( 7 * 15)

However, since each game between two teams is performed only once, each team in a row cannot play with the previous row because they already played a game.

bc can only combine with teams that don't have b or c or a (because combinations with 'a' have already played):
bcde bcdf bcdg bcdh bcef bceg bceh bcfg bcfh bcgh

so each team in the 2nd row can combine with 10 other teams for a game (6 *10)

using that same logic, go down to the next row. cd can only play with teams that don't have c or d or a or b (because teams with a or b have already played):
cdef cdeg cdeh cdfh cdfh cdgh

so each team in the 3rd row can play with 6 other teams for a game (5 *6)

next row is 3 other teams. Check de:
defgh defh degh --> (4 * 3)

next row is 1 other team. check ef:
efgh (3 * 1)


the following rows will have nothing more to play with because any combination with them have already played, so the total number of games will be:

(7*15) + (6*10) + (5*6) +(4*3) + (3*1) = 210

I believe the answer should be 210.


(Although 210 is a factor of 630, I don't see how you can get to 630).