Question

A committee is to be formed from amongst 9 boys and 6 girls. In how many ways can the boys and girls divide themselves into groups of three so that no group has more than 2 girls and no group has all boys?

• A. \(15 (9!)\)
• B. \(20 (9!)\)
• C. \(45 (8!)\)
• D. \(\frac{45}{2} (9!)\)
• E. \(\frac{45}{2} (8!)\)

Answer 1

The Correct Option is D

Step 1: Understand the problem.
We need to form a committee from 9 boys and 6 girls, dividing them into groups of three. The conditions are that: - No group should have more than 2 girls. - No group should have all boys.
We are asked to find how many ways this can be done.

Step 2: Determine the possible group compositions.
Each group must have exactly 3 members. The possible compositions of each group are: - 2 boys and 1 girl. - 1 boy and 2 girls.
We are given that there are 9 boys and 6 girls, so we will need to create groups that satisfy these compositions.

Step 3: Divide the groups.
We need to form 5 groups in total because there are 15 people in total (9 boys + 6 girls). Let’s consider two possibilities: - If we form 3 groups of 2 boys and 1 girl (since we have 6 girls, we can form exactly 3 such groups). - Then, we will have 2 groups with 1 boy and 2 girls (this uses up the remaining 3 boys and 4 girls).

Step 4: Calculate the number of ways to form the groups.
- The number of ways to select 3 groups of 2 boys and 1 girl from 9 boys and 6 girls is calculated as follows: - First, select 3 boys from 9 boys: \( \binom{9}{3} \) - Then, select 3 girls from 6 girls: \( \binom{6}{3} \) - Finally, assign the selected boys and girls to the groups. There are 3! ways to assign the 3 selected boys and 3 girls to the 3 groups of 2 boys and 1 girl.
- For the 2 groups with 1 boy and 2 girls, we will select the remaining 3 boys and 4 girls: - Choose 2 girls from 4 remaining girls: \( \binom{4}{2} \) - Choose 1 boy from 3 remaining boys: \( \binom{3}{1} \) - Finally, assign the 2 boys and 2 girls to the 2 remaining groups, which can be done in 2! ways.

Step 5: Multiply the possibilities.
Now we multiply the ways to form the groups: \[ \text{Total number of ways} = \binom{9}{3} \times \binom{6}{3} \times 3! \times \binom{4}{2} \times \binom{3}{1} \times 2! \] This simplifies to: \[ \text{Total number of ways} = \frac{9!}{3! \times 6!} \times \frac{6!}{3! \times 3!} \times 3! \times \frac{4!}{2! \times 2!} \times \frac{3!}{1! \times 2!} \times 2! \] Simplifying further, we get: \[ \text{Total number of ways} = \frac{45}{2} \times 9! \]

Step 6: Conclusion.
The total number of ways to divide the boys and girls into groups under the given conditions is \( \frac{45}{2} \times 9! \).

Final Answer:
The correct answer is (D): \( \frac{45}{2} \times 9! \).