MATH SOLVE

3 months ago

Q:
# There are 5 boys and 5 girls on a co-ed basketball team. A team plays 5 players on the court at one time, and the team can have at most 3 boys. How many possible teams can be created?

Accepted Solution

A:

Answer:226 teamsStep-by-step explanation:A team can be:5 girls4 girls and 1 boy3 girls and 2 boys2 girls and 3 boysCount all combinations:1. 5 girls from 5 girls can be chosen in 1 way.2. 4 girls from 5 girls can be chosen in [tex]C^5_4=\dfrac{5!}{4!(5-4)!}=\dfrac{1\cdot 2\cdot 3\cdot 4\cdot 5}{1\cdot 2\cdot 3\cdot 4\cdot 1}=5[/tex]different ways and 1 boy from 5 boys can be chosen in [tex]C^5_1=\dfrac{5!}{1!(5-1)!}=\dfrac{5!}{1!\cdot 4!}=\dfrac{1\cdot 2\cdot 3\cdot 4\cdot 5}{1\cdot 2\cdot 3\cdot 4\cdot 1}=5[/tex]different ways, so in total, [tex]5\cdot 5=25[/tex]different combinations.3. 3 girls from 5 girls can be chosen in [tex]C^5_3=\dfrac{5!}{3!(5-3)!}=\dfrac{5!}{3!\cdot 2!}=\dfrac{1\cdot 2\cdot 3\cdot 4\cdot 5}{1\cdot 2\cdot 3\cdot 1\cdot 2}=10[/tex]different ways and 2 boys from 5 boys can be chosen in [tex]C^5_2=\dfrac{5!}{2!(5-2)!}=\dfrac{5!}{2!\cdot 3!}=\dfrac{1\cdot 2\cdot 3\cdot 4\cdot 5}{1\cdot 2\cdot 1\cdot 2\cdot 3}=10[/tex]different ways, so in total, [tex]10\cdot 10=100[/tex]different combinations.4. 2 girls from 5 girls can be chosen in [tex]C^5_2=\dfrac{5!}{2!(5-2)!}=\dfrac{5!}{2!\cdot 3!}=\dfrac{1\cdot 2\cdot 3\cdot 4\cdot 5}{1\cdot 2\cdot 1\cdot 2\cdot 3}=10[/tex]different ways and 3 boys from 5 boys can be chosen in [tex]C^5_3=\dfrac{5!}{3!(5-3)!}=\dfrac{5!}{3!\cdot 2!}=\dfrac{1\cdot 2\cdot 3\cdot 4\cdot 5}{1\cdot 2\cdot 3\cdot 1\cdot 2}=10[/tex]different ways, so in total, [tex]10\cdot 10=100[/tex]different combinations.In total, there are [tex]1+25+100+100=226[/tex] possible teams.