Get 75 points big dawg and awnser this for me and no its not 16. Eight two-person teams are participating in a three-legged race. Find the number of orders in which all 8 teams can finish the race.

Each team needs two people for a 3-legged race.

Therefore, the number of orders (combinations) of 2 people at a time from 8 is

₈C₂ = 8!/(2!6!)

     = (8*7*6!)/(2*6!)

     = (8*7)/2

     = 56/2

     = 28

28

Each team needs two people for a 3-legged race.

Therefore, the number of orders of 2 people at a time from 8 is

₈C₂ = 8!/(2!6!)

    = (8*7*6!)/(2*6!)

    = (8*7)/2

    = 56/2

    = 28

After solving, you will be able to know that your answer is 28

i think is g(1)=-1.

hope that i can

just sum

10 + 10 = 20