A hockey team is made up of 5 players each from Scotland, Wales, and Ireland, and 3 from England. In how many ways can the team of 11 players be selected such that there are more English players than Irish players?
This is a real-life problem I was trying to figure out, and my attempt was to count through every possible selection where there are more English players than Irish players - e.g 3E,1I/3E,2I/2E,1I/3E,0I/2E,0I/1E,0I, however, this is a very tedious process and I was wondering what the best solution to this problem is. All help appreciated. Thank you
Let $S$ be the set of pairs $(e,i)$, where $e,i$ are integers with $1\le e \le 3$ and $0 \le i < e$.
Explicitly we have $S=\{(1,0),(2,0),(2,1),(3,0),(3,1),(3,2)\}$.
For $(e,i)\in S$, the number of ways to form a team with $e$ English players and $i$ Irish players is $$\binom{3}{e}\binom{5}{i}\binom{10}{11-(e+i)}$$ hence the number of teams satisfying the specified condition is $$ {\large{\sum_{(e,i)\in S}}}\binom{3}{e}\binom{5}{i}\binom{10}{11-(e+i)} = 3+30+675+45+600+2100 = 3453 $$