Question

Group A consists of 7 boys and 3 girls, while group B consists of 6 boys and 5 girls. The number of ways, 4 boys and 4 girls can be invited for a picnic if 5 of them must be from group A and the remaining 3 from group B, is equal to:

• A. \( 8925 \)
• B. \( 8750 \)
• C. \( 9100 \)
• D. \( 8575 \)

Hint:

To solve problems involving combinations, break down the selections into manageable cases and then sum the results for all valid combinations.

Answer 1

The Correct Option is B

It is required to select 5 individuals from group A (comprising 7 boys and 3 girls) and 3 individuals from group B (comprising 6 boys and 5 girls).

For group A, the selection can be 4 boys and 1 girl, or 3 boys and 2 girls. The number of ways to achieve these selections is calculated using combinations:

\[ \text{Ways for group A} = \binom{7}{4} \times \binom{3}{1} + \binom{7}{3} \times \binom{3}{2}. \]

For group B, the remaining individuals are selected as follows:

\[ \text{Ways for group B} = \binom{6}{1} \times \binom{5}{2} + \binom{6}{2} \times \binom{5}{1}. \]

The total number of ways for both groups is found by multiplying the respective ways.

Final Answer: 8750.