Key Concepts and Formulas

• Combinations: The number of ways to choose $r$ items from a set of $n$ distinct items, where order doesn't matter, is given by the combination formula: $\binom{n}{r} = \frac{n!}{r!(n-r)!}$
• Complementary Counting: A counting technique where you find the total number of outcomes and subtract the number of unfavorable outcomes to find the number of favorable outcomes.
• Multiplication Principle: If there are $m$ ways to do one thing and $n$ ways to do another, then there are $m \times n$ ways to do both.

Step-by-Step Solution

1. Calculate the Total Number of Possible Teams Without Restrictions

First, we'll find the total number of ways to form a team of 2 girls and 3 boys from a class of 5 girls and 7 boys, without considering the restriction on boys A and B.

• Selecting Girls: We need to choose 2 girls out of 5.

  • Why: The order of selection doesn't matter, so it's a combination.
  • Calculation: $\binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5!}{2!3!} = \frac{5 \times 4}{2 \times 1} = 10 \text{ ways}$

• Selecting Boys: We need to choose 3 boys out of 7.

  • Why: Again, order doesn't matter, so it's a combination.
  • Calculation: $\binom{7}{3} = \frac{7!}{3!(7-3)!} = \frac{7!}{3!4!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35 \text{ ways}$

• Total Unrestricted Teams: To get the total number of teams without restrictions, we multiply the number of ways to choose the girls by the number of ways to choose the boys.

  • Why: By the multiplication principle, each choice of girls can be combined with each choice of boys.
  • Calculation: $\text{Total unrestricted teams} = \binom{5}{2} \times \binom{7}{3} = 10 \times 35 = 350 \text{ teams}$

2. Calculate the Number of Teams Where Both A and B are Included

Now, we'll calculate the number of teams that violate the condition, i.e., teams that include both boys A and B.

• Selecting Girls: We still need to choose 2 girls out of 5. The presence of A and B doesn't affect this selection.

  • Why: The condition only applies to the boys.
  • Calculation: $\binom{5}{2} = 10 \text{ ways}$

• Selecting Boys (with A and B included): Since A and B must be on the team, we only need to choose 1 more boy to complete the team of 3.

  • Why: A and B occupy two of the three slots for boys.
  • Remaining Boys to Choose: We need to choose $3 - 2 = 1$ boy.
  • Number of Boys Available: Since A and B are already chosen, we have $7 - 2 = 5$ boys to choose from.
  • Calculation: $\binom{5}{1} = \frac{5!}{1!(5-1)!} = \frac{5!}{1!4!} = 5 \text{ ways}$

• Teams with A and B Together: To get the number of teams where both A and B are present, we multiply the number of ways to choose the girls by the number of ways to choose the remaining boy (given A and B are already selected).

  • Why: The multiplication principle.
  • Calculation: $\text{Teams with A and B together} = \binom{5}{2} \times \binom{5}{1} = 10 \times 5 = 50 \text{ teams}$

3. Calculate the Number of Teams That Satisfy the Condition (A and B are not Both Together)

Finally, we use complementary counting. We subtract the number of "unfavorable" teams (where A and B are together) from the total number of unrestricted teams.

• Why: This gives us the number of teams where A and B are not both on the team.
• Calculation: $\text{Required number of teams} = \text{Total unrestricted teams} - \text{Teams with A and B together}$ $\text{Required number of teams} = 350 - 50 = 300 \text{ teams}$

Common Mistakes & Tips

• Misinterpreting the restriction: The problem states A and B refuse to be both on the same team. This does not mean they can never be on a team at all.
• Forgetting to adjust the pool after forcing members: When you force A and B onto the team, you must reduce the number of boys you still need to choose and the total number of boys available.
• Choosing the right strategy: Complementary counting is often the easiest approach when dealing with "not together," "at least," or "at most" type problems.

Summary

This problem demonstrates the use of combinations and complementary counting to solve a seemingly complex counting problem. By calculating the total number of possible teams and then subtracting the number of teams that violate the given condition, we arrive at the solution. The number of different teams consisting of 2 girls and 3 boys that can be formed, given that boys A and B refuse to be members of the same team, is 300.

The final answer is $\boxed{300}$, which corresponds to option (D).