Team Selection C11150

1. **Problem Statement:** We have $10$ students with the following skills: - $5$ can install computers. - $3$ can install audio equipment. - $2$ can install both computers and audio equipment. We want to select a team of $6$ people such that: - $3$ people can install computers. - $3$ people can install audio equipment. We need to find the number of different selection schemes. 2. **Understanding the problem:** Let $x$ be the number of people who can install both computers and audio equipment selected in the team. Since $3$ people are required for computers and $3$ for audio equipment, and $x$ people can do both, the total team size is $6$. 3. **Constraints on $x$:** - $x$ must be at least $0$. - $x$ cannot exceed the number of people who can do both, which is $2$. - Also, the number of people selected only for computers is $3 - x$. - The number of people selected only for audio equipment is $3 - x$. 4. **Available people:** - People who can only install computers: $5 - 2 = 3$. - People who can only install audio equipment: $3 - 2 = 1$. - People who can do both: $2$. 5. **Calculate the number of ways for each possible $x$:** - For $x = 0$: - Select $3$ from $3$ computer-only people: $\binom{3}{3} = 1$ - Select $3$ from $1$ audio-only people: $\binom{1}{3} = 0$ (impossible) - Select $0$ from $2$ both-skilled people: $\binom{2}{0} = 1$ - Total ways: $1 \times 0 \times 1 = 0$ - For $x = 1$: - Select $2$ from $3$ computer-only people: $\binom{3}{2} = 3$ - Select $2$ from $1$ audio-only people: $\binom{1}{2} = 0$ (impossible) - Select $1$ from $2$ both-skilled people: $\binom{2}{1} = 2$ - Total ways: $3 \times 0 \times 2 = 0$ - For $x = 2$: - Select $1$ from $3$ computer-only people: $\binom{3}{1} = 3$ - Select $1$ from $1$ audio-only people: $\binom{1}{1} = 1$ - Select $2$ from $2$ both-skilled people: $\binom{2}{2} = 1$ - Total ways: $3 \times 1 \times 1 = 3$ 6. **Sum all valid ways:** $$\text{Total ways} = 0 + 0 + 3 = 3$$ **Final answer:** There are $3$ different selection schemes.