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

From group A, with 7 boys and 3 girls, 5 individuals must be selected. From group B, with 6 boys and 5 girls, 3 individuals must be selected. The selection from group A can be done by choosing 3 boys and 2 girls, or 4 boys and 1 girl. The number of ways to achieve this is calculated as: \[\text{Ways for group A} = \binom{7}{4} \times \binom{3}{1} + \binom{7}{3} \times \binom{3}{2}.\] The selection from group B involves choosing the remaining individuals 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 is obtained by multiplying the ways for each group. Final Answer: 8750.