Key Concepts and Formulas

• Combinations ($\binom{n}{r}$): The number of ways to choose $r$ items from a set of $n$ distinct items without regard to order: $\binom{n}{r} = \frac{n!}{r!(n-r)!}$.
• Principle of Complementary Counting: The number of ways an event A occurs is the total number of ways minus the number of ways A does not occur: $N(A) = \text{Total} - N(A')$.
• 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: Calculate the total number of ways to form a group of 3 boys and 3 girls without any restrictions.

We first find the number of ways to select 3 boys out of 10 and 3 girls out of 5. Since the order of selection doesn't matter, we use combinations.

• Ways to select 3 boys from 10: $\binom{10}{3} = \frac{10!}{3!(10-3)!} = \frac{10!}{3!7!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 10 \times 3 \times 4 = 120$
• Ways to select 3 girls from 5: $\binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5!}{3!2!} = \frac{5 \times 4}{2 \times 1} = 10$

Now, we use the multiplication principle to find the total number of ways to form the group:

• Total number of groups (without restrictions): $\binom{10}{3} \times \binom{5}{3} = 120 \times 10 = 1200$

Step 2: Calculate the number of ways to form a group where both B1 and B2 are members.

If both B1 and B2 are in the group, we need to choose one more boy from the remaining 8 boys (B3, B4, ..., B10). We still need to choose 3 girls from the 5 available.

• Ways to select the remaining 1 boy from 8: $\binom{8}{1} = \frac{8!}{1!(8-1)!} = \frac{8!}{1!7!} = 8$
• Ways to select 3 girls from 5 (same as before): $\binom{5}{3} = 10$

Now, we use the multiplication principle to find the total number of groups with both B1 and B2:

• Total number of groups with B1 and B2: $\binom{8}{1} \times \binom{5}{3} = 8 \times 10 = 80$

Step 3: Apply the Principle of Complementary Counting.

We subtract the number of groups where B1 and B2 are together from the total number of possible groups to find the number of groups where B1 and B2 are not both members.

• Number of groups where B1 and B2 are not both members: $1200 - 80 = 1120$

Common Mistakes & Tips

• Ensure you are using combinations (order doesn't matter) and not permutations (order matters).
• Remember to adjust the number of boys to choose from when B1 and B2 are already selected.
• The principle of complementary counting is useful when directly calculating the desired outcome is difficult.

Summary

We found the total number of ways to form a group of 3 boys and 3 girls without any restrictions, then subtracted the number of groups where both B1 and B2 are members to find the number of groups where they are not both members. This gives us the desired number of groups as 1120.

The final answer is \boxed{1120}.