Question

A man has 9 friends: 4 boys and 5 girls. In how many ways can he invite them, if there have to be exactly 3 girls in the invitees?

• A. \( \binom{5}{3} \times \binom{4}{2} \)
• B. \( \binom{5}{3} \times \binom{4}{3} \)
• C. \( \binom{5}{3} \times \binom{4}{4} \)
• D. \( \binom{5}{3} \times \binom{4}{1} \)

Hint:

Use the combination formula \( \binom{n}{r} \) to calculate the number of ways to select items without regard to order.

Answer 1

The Correct Option is A

We are given 9 friends, 4 boys and 5 girls, and the condition that exactly 3 girls must be invited. Step 1: First, select the 3 girls from the 5 girls. The number of ways to do this is given by: \[ \binom{5}{3} = \frac{5 \times 4 \times 3}{3 \times 2 \times 1} = 10 \] Step 2: After selecting the 3 girls, we need to select 2 boys from the 4 boys. The number of ways to do this is given by: \[ \binom{4}{2} = \frac{4 \times 3}{2 \times 1} = 6 \] Step 3: Multiply the results of Step 1 and Step 2 to get the total number of ways to invite 3 girls and 2 boys: \[ \binom{5}{3} \times \binom{4}{2} = 10 \times 6 = 60 \] Thus, the answer is: 60.