The baseball team is made up of 7 girls and 8 boys. There are 9 people on the field at the same time in a baseball game. If at least four of the people on the field must be girls, how many different possible ways can the team take the field?
So far all I've managed to try was 15C9, and then divided that by 3! to remove possible combinations where there are less than 4 girls. This results in a decimal answer, though.
Any help is appreciated!
Best Answer
there're 4 different situation about the 9-people team: 4-girl, 5-girl, 6-girl, and 7-girl. let's do it one by one. note that if there're 4 girls, there must be 5 boys.
4-girl: $C^4_7 \cdot C^5_8 = 35 \cdot 56 = 1960$
5-girl: $C^5_7 \cdot C^4_8 = 21 \cdot 70 = 1470$
6-girl: $C^6_7 \cdot C^3_8 = 7 \cdot 56 = 392$
7-girl: $C^7_7 \ cdot C^2_8 = 1 \cdot 28 = 28$
so there're $1960 + 1470 + 392 + 28 = 3850$ combinations.