Key Concepts and Formulas

• Combinations: The number of ways to choose $r$ objects from a set of $n$ distinct objects is given by the combination formula: ${n \choose r} = \frac{n!}{r!(n-r)!}$, where $n!$ (n factorial) is the product of all positive integers up to $n$.
• 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

Step 1: Understand the problem

We need to select 4 boys and 4 girls for a picnic such that 5 people are from group A and 3 are from group B. We need to find the number of ways this can be done.

Step 2: Identify the possible cases

Since 5 people must be from group A and 3 from group B, we need to consider all the possible combinations of boys and girls that satisfy this condition, given the compositions of group A (7 boys, 3 girls) and group B (6 boys, 5 girls). The total number of boys selected must be 4, and the total number of girls selected must be 4. Let's denote the number of girls from group A as $g_A$ and the number of boys from group A as $b_A$. Similarly, $g_B$ and $b_B$ represent the number of girls and boys from group B, respectively. We have:

• $g_A + b_A = 5$
• $g_B + b_B = 3$
• $g_A + g_B = 4$
• $b_A + b_B = 4$

We need to find integer solutions to these equations, keeping in mind the constraints $0 \le g_A \le 3$, $0 \le b_A \le 7$, $0 \le g_B \le 5$, and $0 \le b_B \le 6$.

Let's consider the possible values for $g_A$.

• If $g_A = 1$, then $b_A = 5$, $g_B = 3$, $b_B = 0$
• If $g_A = 2$, then $b_A = 3$, $g_B = 2$, $b_B = 1$
• If $g_A = 3$, then $b_A = 2$, $g_B = 1$, $b_B = 2$

These are the only three possible cases:

• Case I: 3 girls from A and 2 boys from A, and 1 girl from B and 2 boys from B. $(g_A=3, b_A=2, g_B=1, b_B=2)$
• Case II: 2 girls from A and 3 boys from A, and 2 girls from B and 1 boy from B. $(g_A=2, b_A=3, g_B=2, b_B=1)$
• Case III: 1 girl from A and 4 boys from A, and 3 girls from B and 0 boys from B. $(g_A=1, b_A=4, g_B=3, b_B=0)$

Step 3: Calculate the number of ways for each case

• Case I: Selecting 3 girls from A and 2 boys from A, and 1 girl from B and 2 boys from B. Number of ways = ${3 \choose 3} \times {7 \choose 2} \times {5 \choose 1} \times {6 \choose 2} = 1 \times \frac{7 \times 6}{2 \times 1} \times 5 \times \frac{6 \times 5}{2 \times 1} = 1 \times 21 \times 5 \times 15 = 1575$

• Case II: Selecting 2 girls from A and 3 boys from A, and 2 girls from B and 1 boy from B. Number of ways = ${3 \choose 2} \times {7 \choose 3} \times {5 \choose 2} \times {6 \choose 1} = 3 \times \frac{7 \times 6 \times 5}{3 \times 2 \times 1} \times \frac{5 \times 4}{2 \times 1} \times 6 = 3 \times 35 \times 10 \times 6 = 6300$

• Case III: Selecting 1 girl from A and 4 boys from A, and 3 girls from B and 0 boys from B. Number of ways = ${3 \choose 1} \times {7 \choose 4} \times {5 \choose 3} \times {6 \choose 0} = 3 \times \frac{7 \times 6 \times 5}{3 \times 2 \times 1} \times \frac{5 \times 4}{2 \times 1} \times 1 = 3 \times 35 \times 10 \times 1 = 1050$

Step 4: Calculate the total number of ways

Total number of ways = Number of ways in Case I + Number of ways in Case II + Number of ways in Case III Total number of ways = $1575 + 6300 + 1050 = 8925$

Common Mistakes & Tips

• Missing a case: Ensure all possible combinations of boys and girls are considered. A systematic approach helps avoid this.
• Incorrect application of combinations: Double-check the values of $n$ and $r$ in the ${n \choose r}$ formula.
• Not applying the multiplication principle: Remember to multiply the number of ways to choose boys and girls from each group.

Summary

The problem involves selecting 4 boys and 4 girls from two groups with specific compositions, with the constraint that 5 people come from group A and 3 from group B. We identified three possible cases based on the number of girls selected from group A. For each case, we calculated the number of ways to select the required number of boys and girls from each group using combinations. Finally, we summed the number of ways for each case to find the total number of ways to form the picnic group, which is 8925.

Final Answer

The final answer is \boxed{8925}, which corresponds to option (A).