Hmmm, my math skills are a little weak, so maybe somebody can do the work for me. With 11 starters, and say 10 substitutes. What is the possible number of player combinations that can be used.
"TheEngineer" wrote:
It's a bit more complicated than you'd might think.
Firstly you need to establish what positions the substitutes play.
Let's start with an easy example:
3 starters on a team, each person has a backup.
Let a denote the first person, b denote the second, c denote the third, d = a's backup, e = b's backup, f = c's backup.
So we could have starters from a pool of abc, abf, aec, dbc, dec, dbf, aef, def = 8 possibilities.
This is a pool of 6 in total number where we choose three 3 starters from a binomial. Hence 2^3 = 8.
For 20 players, e.g. 10 positions with 2 players each, from a pool of 20 selections, we get 2^10 = 1024 (the lone starter always remains a starter and hence doesn't affect the number of combinations).
Since you can blitz any number of people, for a given starting 11, you can blitz from 1 up to 11 people. This is the sum of 11C1 + 11C2 ... up to 11C11 = 2047 blitzing combinations per starting 11. Since there are 1024 possible starting 11s, 1024 * 2047 = 2,096,128 possible blitzing combinations.
"PackFanWithTwins" wrote: